Question 82: The Finite Expansion Includes Every k from Zero through n

Original question and givens

Use natural exponent n=6, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary).

• G001 — Use natural exponent n=6, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary).

Definition and scope

n is a nonnegative integer and x,y commute. For n=0 use the polynomial empty-product identity.

• mko-binomial-theorem

Step-by-step derivation

n=6≥0; x,y commute

(x+y)^6 is a finite product of 6 factors

choose exactly 0 of 6 factors to contribute y

C(6,0)=720/(1·720)=1

C(6,0)=1; choose y 0 times

1·x^6·y^0

choose exactly 1 of 6 factors to contribute y

C(6,1)=720/(1·120)=6

C(6,1)=6; choose y 1 times

6·x^5·y^1

choose exactly 2 of 6 factors to contribute y

C(6,2)=720/(2·24)=15

C(6,2)=15; choose y 2 times

15·x^4·y^2

choose exactly 3 of 6 factors to contribute y

C(6,3)=720/(6·6)=20

C(6,3)=20; choose y 3 times

20·x^3·y^3

choose exactly 4 of 6 factors to contribute y

C(6,4)=720/(24·2)=15

C(6,4)=15; choose y 4 times

15·x^2·y^4

choose exactly 5 of 6 factors to contribute y

C(6,5)=720/(120·1)=6

C(6,5)=6; choose y 5 times

6·x^1·y^5

choose exactly 6 of 6 factors to contribute y

C(6,6)=720/(720·1)=1

C(6,6)=1; choose y 6 times

1·x^0·y^6

all k=0 through 6 have been accounted for

(x+y)^6=1·x^6·y^0 + 6·x^5·y^1 + 15·x^4·y^2 + 20·x^3·y^3 + 15·x^2·y^4 + 6·x^1·y^5 + 1·x^0·y^6

Reverse checks

• S016 · premise_and_result_audit — There are n+1 coefficient positions, with the complete range of k retained.
• S015 · premise_and_result_audit — The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns.
• S014 · premise_and_result_audit — Check the coefficient independently with the integer Pascal recurrence, not approximate factorials.
• S013 · premise_and_result_audit — The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns.
• S012 · premise_and_result_audit — Check the coefficient independently with the integer Pascal recurrence, not approximate factorials.
• S011 · premise_and_result_audit — The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns.
• S010 · premise_and_result_audit — Check the coefficient independently with the integer Pascal recurrence, not approximate factorials.
• S009 · premise_and_result_audit — The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns.
• S008 · premise_and_result_audit — Check the coefficient independently with the integer Pascal recurrence, not approximate factorials.
• S007 · premise_and_result_audit — The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns.
• S006 · premise_and_result_audit — Check the coefficient independently with the integer Pascal recurrence, not approximate factorials.
• S005 · premise_and_result_audit — The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns.
• S004 · premise_and_result_audit — Check the coefficient independently with the integer Pascal recurrence, not approximate factorials.
• S003 · premise_and_result_audit — The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns.
• S002 · premise_and_result_audit — Check the coefficient independently with the integer Pascal recurrence, not approximate factorials.
• S001 · premise_and_result_audit — This is not an infinite expansion for a noninteger power or a claim for noncommuting multiplication.

Result

(6, [[0, 1, 6, 0], [1, 6, 5, 1], [2, 15, 4, 2], [3, 20, 3, 3], [4, 15, 2, 4], [5, 6, 1, 5], [6, 1, 0, 6]], complete_k_zero_through_n_coefficients)

The full expansion consists of all displayed monomials, in the same order as the answer rows.

This question review and source binding

Review date: 2026-10-24

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.