The coefficient of x^{37} in the expansion of | vedclass

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Question

(C) Given expression: $(1-x)^{30} (1+x+x^2)^{29}$.We can rewrite this as $(1-x) (1-x)^{29} (1+x+x^2)^{29}$.Since $(1-x)(1+x+x^2) = (1-x^3)$,the expres

Vedclass · Accepted Answer

$-{}^{29}C_{12}$

Vedclass · Answer

(C) Given expression: $(1-x)^{30} (1+x+x^2)^{29}$.We can rewrite this as $(1-x) (1-x)^{29} (1+x+x^2)^{29}$.Since $(1-x)(1+x+x^2) = (1-x^3)$,the expression becomes $(1-x) (1-x^3)^{29}$.Expanding this: $(1-x^3)^{29} - x(1-x^3)^{29}$.We need the coefficient of $x^{37}$.In the first part $(1-x^3)^{29}$,the powers of $x$ are of the form $x^{3k}$. Since $37$ is not divisible by $3$,the coefficient of $x^{37}$ is $0$.In the second part $-x(1-x^3)^{29}$,we need the coefficient of $x^{36}$ in $(1-x^3)^{29}$.Let $x^3 = y$,then we need the coefficient of $y^{12}$ in $(1-y)^{29}$.The general term is $T_{r+1} = {}^{29}C_r (-y)^r$.For $y^{12}$,$r=12$,so the term is ${}^{29}C_{12} (-y)^{12} = {}^{29}C_{12} y^{12}$.Including the $-x$ factor,the term is $-x({}^{29}C_{12} x^{36}) = -{}^{29}C_{12} x^{37}$.Thus,the coefficient is $-{}^{29}C_{12}$.

The coefficient of $x^{37}$ in the expansion of $(1-x)^{30} (1 + x + x^2)^{29}$ is:

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