$(1+x+x^2+cdots+x^12)^3 = 1+cdots+ Ax^12$ find A =?

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Answer 1 · 2018-08-12T06:04:05

$91$

$1 + x + x^2 + ... + x^12 = (1 - x^13)/(1 - x)$
$=> (1 + x + x^2 + ... + x^12)^3 = (1 - x^13)^3 / (1-x)^3$

$1/(1 - x)^3 = (1 + x + x^2 + x^3 + ... )^3$
$= 1 + C(3,1) x + C(4,2) x^2 + C(5,3) x^3 + ...$

$"So we have"$

$(1 - 3 x^13 + 3 x^26 - x^39)(1 + 3 x + 6 x^2 + 10 x^3 + ...)$

$"We need the coefficient of "x^12.$

$"This coefficient is "C(14,12) = 91.$

(1+x+x^2+cdots+x^12)^3 = 1+cdots+ Ax^12(1+x+x^2+cdots+x^12)^3 = 1+cdots+ Ax^12 find A =?