# Finite Induction Scope and Separate Base/Step Defects This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed. This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately. batch-20261008-advance-20261012-r01-mathematical-induction-foundations · 2026-10-08 · 100 questions This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt. ## 1. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=0. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(0, 1) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (0, 1). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0001/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 2. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=1. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(1, 2) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (1, 2). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0002/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 3. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=2. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(2, 3) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (2, 3). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0003/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 4. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=3. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(3, 4) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (3, 4). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0004/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 5. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=4. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(4, 5) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (4, 5). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0005/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 6. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=5. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(5, 6) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (5, 6). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0006/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 7. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=6. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(6, 7) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (6, 7). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0007/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 8. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=7. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(7, 8) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (7, 8). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0008/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 9. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=8. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(8, 9) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (8, 9). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0009/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 10. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=9. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(9, 10) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (9, 10). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0010/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 11. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=10. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(10, 11) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (10, 11). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0011/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 12. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=11. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(11, 12) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (11, 12). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0012/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 13. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=12. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(12, 13) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (12, 13). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0013/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 14. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=13. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(13, 14) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (13, 14). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0014/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 15. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=14. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(14, 15) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (14, 15). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0015/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 16. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=15. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(15, 16) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (15, 16). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0016/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 17. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=16. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(16, 17) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (16, 17). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0017/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 18. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=17. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(17, 18) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (17, 18). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0018/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 19. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=18. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(18, 19) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (18, 19). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0019/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 20. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=19. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(19, 20) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (19, 20). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0020/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 21. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=20. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(20, 21) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (20, 21). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0021/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 22. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=21. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(21, 22) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (21, 22). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0022/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 23. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=22. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(22, 23) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (22, 23). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0023/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 24. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=23. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(23, 24) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (23, 24). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0024/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 25. Induction-Hypothesis and Successor Indices N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. n=24. Give (index n of hypothesis P, index n+1 of the successor target), without claiming P holds. View answer and explanation(24, 25) Compute the successor of the nonnegative index; identifying scope does not prove the proposition. The exact result, retaining the requested order, is (24, 25). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0025/) Mathematical induction · qf-density12-induction-successor-index-pair · candidate ## 26. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-2, a(k+1)=a(k)+(99), through=0. List a(0) through a(through), inclusive. View answer and explanation[-2] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-2]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0026/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 27. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-11, a(k+1)=a(k)+(-1), through=2. List a(0) through a(through), inclusive. View answer and explanation[-11, -12, -13] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-11, -12, -13]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0027/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 28. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-10, a(k+1)=a(k)+(0), through=3. List a(0) through a(through), inclusive. View answer and explanation[-10, -10, -10, -10] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-10, -10, -10, -10]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0028/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 29. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-9, a(k+1)=a(k)+(1), through=4. List a(0) through a(through), inclusive. View answer and explanation[-9, -8, -7, -6, -5] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-9, -8, -7, -6, -5]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0029/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 30. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-8, a(k+1)=a(k)+(2), through=5. List a(0) through a(through), inclusive. View answer and explanation[-8, -6, -4, -2, 0, 2] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-8, -6, -4, -2, 0, 2]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0030/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 31. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-7, a(k+1)=a(k)+(-2), through=6. List a(0) through a(through), inclusive. View answer and explanation[-7, -9, -11, -13, -15, -17, -19] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-7, -9, -11, -13, -15, -17, -19]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0031/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 32. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-6, a(k+1)=a(k)+(-1), through=7. List a(0) through a(through), inclusive. View answer and explanation[-6, -7, -8, -9, -10, -11, -12, -13] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-6, -7, -8, -9, -10, -11, -12, -13]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0032/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 33. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-5, a(k+1)=a(k)+(0), through=8. List a(0) through a(through), inclusive. View answer and explanation[-5, -5, -5, -5, -5, -5, -5, -5, -5] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-5, -5, -5, -5, -5, -5, -5, -5, -5]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0033/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 34. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-4, a(k+1)=a(k)+(1), through=1. List a(0) through a(through), inclusive. View answer and explanation[-4, -3] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-4, -3]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0034/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 35. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-3, a(k+1)=a(k)+(2), through=2. List a(0) through a(through), inclusive. View answer and explanation[-3, -1, 1] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-3, -1, 1]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0035/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 36. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-2, a(k+1)=a(k)+(-2), through=3. List a(0) through a(through), inclusive. View answer and explanation[-2, -4, -6, -8] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-2, -4, -6, -8]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0036/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 37. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=-1, a(k+1)=a(k)+(-1), through=4. List a(0) through a(through), inclusive. View answer and explanation[-1, -2, -3, -4, -5] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [-1, -2, -3, -4, -5]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0037/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 38. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=0, a(k+1)=a(k)+(0), through=5. List a(0) through a(through), inclusive. View answer and explanation[0, 0, 0, 0, 0, 0] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [0, 0, 0, 0, 0, 0]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0038/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 39. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=1, a(k+1)=a(k)+(1), through=6. List a(0) through a(through), inclusive. View answer and explanation[1, 2, 3, 4, 5, 6, 7] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [1, 2, 3, 4, 5, 6, 7]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0039/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 40. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=2, a(k+1)=a(k)+(2), through=7. List a(0) through a(through), inclusive. View answer and explanation[2, 4, 6, 8, 10, 12, 14, 16] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [2, 4, 6, 8, 10, 12, 14, 16]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0040/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 41. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=3, a(k+1)=a(k)+(-2), through=8. List a(0) through a(through), inclusive. View answer and explanation[3, 1, -1, -3, -5, -7, -9, -11, -13] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [3, 1, -1, -3, -5, -7, -9, -11, -13]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0041/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 42. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=4, a(k+1)=a(k)+(-1), through=1. List a(0) through a(through), inclusive. View answer and explanation[4, 3] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [4, 3]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0042/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 43. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=5, a(k+1)=a(k)+(0), through=2. List a(0) through a(through), inclusive. View answer and explanation[5, 5, 5] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [5, 5, 5]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0043/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 44. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=6, a(k+1)=a(k)+(1), through=3. List a(0) through a(through), inclusive. View answer and explanation[6, 7, 8, 9] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [6, 7, 8, 9]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0044/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 45. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=7, a(k+1)=a(k)+(2), through=4. List a(0) through a(through), inclusive. View answer and explanation[7, 9, 11, 13, 15] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [7, 9, 11, 13, 15]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0045/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 46. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=8, a(k+1)=a(k)+(-2), through=5. List a(0) through a(through), inclusive. View answer and explanation[8, 6, 4, 2, 0, -2] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [8, 6, 4, 2, 0, -2]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0046/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 47. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=9, a(k+1)=a(k)+(-1), through=6. List a(0) through a(through), inclusive. View answer and explanation[9, 8, 7, 6, 5, 4, 3] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [9, 8, 7, 6, 5, 4, 3]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0047/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 48. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=10, a(k+1)=a(k)+(0), through=7. List a(0) through a(through), inclusive. View answer and explanation[10, 10, 10, 10, 10, 10, 10, 10] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [10, 10, 10, 10, 10, 10, 10, 10]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0048/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 49. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=11, a(k+1)=a(k)+(1), through=8. List a(0) through a(through), inclusive. View answer and explanation[11, 12, 13, 14, 15, 16, 17, 18, 19] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [11, 12, 13, 14, 15, 16, 17, 18, 19]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0049/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 50. Inclusive Finite Recurrence Prefix N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. a(0)=12, a(k+1)=a(k)+(2), through=1. List a(0) through a(through), inclusive. View answer and explanation[12, 14] Start at the given initial value and add d repeatedly; through=0 lists only the initial value. Natural indices do not require nonnegative sequence values. The exact result, retaining the requested order, is [12, 14]. [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0050/) Mathematical induction · qf-density12-induction-affine-recursive-prefix · candidate ## 51. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=3, increment d=2; candidate c(k)=(2)k+(3), n=0. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(0, 0) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (0, 0). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0051/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 52. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=3, increment d=2; candidate c(k)=(2)k+(1), n=5. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(-2, 0) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (-2, 0). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0052/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 53. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=1, increment d=2; candidate c(k)=(3)k+(1), n=2. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(0, 1) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (0, 1). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0053/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 54. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=1, increment d=-1; candidate c(k)=(-10)k+(5), n=3. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(4, -9) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (4, -9). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0054/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 55. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-2, increment d=0; candidate c(k)=(-8)k+(3), n=4. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(5, -8) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (5, -8). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0055/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 56. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-1, increment d=1; candidate c(k)=(-6)k+(5), n=5. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(6, -7) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (6, -7). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0056/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 57. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=0, increment d=-1; candidate c(k)=(-7)k+(7), n=0. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(7, -6) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (7, -6). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0057/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 58. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=1, increment d=0; candidate c(k)=(-5)k+(9), n=1. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(8, -5) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (8, -5). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0058/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 59. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-2, increment d=1; candidate c(k)=(-3)k+(7), n=2. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(9, -4) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (9, -4). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0059/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 60. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-1, increment d=-1; candidate c(k)=(-4)k+(9), n=3. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(10, -3) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (10, -3). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0060/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 61. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=0, increment d=0; candidate c(k)=(-2)k+(11), n=4. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(11, -2) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (11, -2). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0061/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 62. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=1, increment d=1; candidate c(k)=(0)k+(13), n=5. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(12, -1) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (12, -1). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0062/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 63. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-2, increment d=-1; candidate c(k)=(-1)k+(11), n=0. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(13, 0) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (13, 0). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0063/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 64. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-1, increment d=0; candidate c(k)=(1)k+(13), n=1. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(14, 1) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (14, 1). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0064/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 65. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=0, increment d=1; candidate c(k)=(3)k+(15), n=2. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(15, 2) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (15, 2). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0065/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 66. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=1, increment d=-1; candidate c(k)=(2)k+(17), n=3. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(16, 3) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (16, 3). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0066/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 67. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-2, increment d=0; candidate c(k)=(4)k+(15), n=4. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(17, 4) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (17, 4). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0067/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 68. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-1, increment d=1; candidate c(k)=(6)k+(17), n=5. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(18, 5) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (18, 5). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0068/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 69. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=0, increment d=-1; candidate c(k)=(5)k+(19), n=0. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(19, 6) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (19, 6). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0069/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 70. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=1, increment d=0; candidate c(k)=(7)k+(21), n=1. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(20, 7) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (20, 7). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0070/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 71. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-2, increment d=1; candidate c(k)=(9)k+(19), n=2. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(21, 8) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (21, 8). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0071/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 72. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-1, increment d=-1; candidate c(k)=(8)k+(21), n=3. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(22, 9) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (22, 9). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0072/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 73. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=0, increment d=0; candidate c(k)=(10)k+(23), n=4. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(23, 10) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (23, 10). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0073/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 74. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=1, increment d=1; candidate c(k)=(12)k+(25), n=5. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(24, 11) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (24, 11). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0074/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 75. Separate Base and Successor-Step Defects N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. Required initial value s=-2, increment d=-1; candidate c(k)=(11)k+(23), n=0. Give signed integers (c(0)−s, c(n+1)−c(n)−d). View answer and explanation(25, 12) Substitute at the base and adjacent indices separately; a base defect and step defect are different obligations, and finite checks are not universal induction. The exact result, retaining the requested order, is (25, 12). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0075/) Mathematical induction · qf-density12-induction-candidate-step-defect · candidate ## 76. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(0) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(1, 0, 1) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (1, 0, 1). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0076/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 77. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(1000) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(1001, 1000, 1001) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (1001, 1000, 1001). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0077/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 78. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(1) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(2, 1, 2) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (2, 1, 2). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0078/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 79. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(2) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(3, 2, 3) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (3, 2, 3). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0079/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 80. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(3) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(4, 3, 4) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (4, 3, 4). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0080/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 81. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(4) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(5, 4, 5) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (5, 4, 5). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0081/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 82. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(5) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(6, 5, 6) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (6, 5, 6). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0082/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 83. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(6) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(7, 6, 7) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (7, 6, 7). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0083/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 84. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(7) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(8, 7, 8) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (8, 7, 8). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0084/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 85. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(8) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(9, 8, 9) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (9, 8, 9). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0085/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 86. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(9) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(10, 9, 10) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (10, 9, 10). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0086/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 87. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(10) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(11, 10, 11) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (11, 10, 11). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0087/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 88. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(11) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(12, 11, 12) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (12, 11, 12). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0088/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 89. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(12) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(13, 12, 13) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (13, 12, 13). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0089/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 90. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(13) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(14, 13, 14) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (14, 13, 14). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0090/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 91. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(14) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(15, 14, 15) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (15, 14, 15). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0091/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 92. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(15) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(16, 15, 16) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (16, 15, 16). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0092/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 93. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(16) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(17, 16, 17) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (17, 16, 17). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0093/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 94. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(17) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(18, 17, 18) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (18, 17, 18). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0094/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 95. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(18) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(19, 18, 19) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (19, 18, 19). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0095/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 96. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(19) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(20, 19, 20) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (20, 19, 20). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0096/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 97. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(20) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(21, 20, 21) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (21, 20, 21). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0097/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 98. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(21) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(22, 21, 22) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (22, 21, 22). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0098/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 99. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(22) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(23, 22, 23) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (23, 22, 23). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0099/) Mathematical induction · qf-density12-induction-range-obligations · candidate ## 100. Finite Coverage and Next Uncovered Index N is indexed from 0; this is an explicitly finite induction protocol, not a universal proof. P(0) through P(23) have been checked individually. Give (number of covered P indices, number of adjacent steps, next uncovered target index). View answer and explanation(24, 23, 24) Inclusive zero-based coverage has one more index than its last index; adjacent steps equal that last index, and later indices remain unproved. The exact result, retaining the requested order, is (24, 23, 24). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261008-advance-20261012-r01-mathematical-induction-0100/) Mathematical induction · qf-density12-induction-range-obligations · candidate [Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261008-advance-20261012-r01-mathematical-induction-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261008-advance-20261012-r01-mathematical-induction-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261008-advance-20261012-r01-mathematical-induction-foundations/validation-report.json)