Finite Induction Scope and Separate Base/Step Defects
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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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Mathematical induction · qf-density12-induction-range-obligations · candidate
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