The coefficient of x^{24} in the expansion of | vedclass

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(B) We are given the expression $(1+x^2)^{12}(1+x^{12})(1+x^{24})$.First,expand the product $(1+x^{12})(1+x^{24}) = 1 + x^{12} + x^{24} + x^{36}$.Now,

Vedclass · Accepted Answer

${}^{12}C_6+2$

Vedclass · Answer

(B) We are given the expression $(1+x^2)^{12}(1+x^{12})(1+x^{24})$.First,expand the product $(1+x^{12})(1+x^{24}) = 1 + x^{12} + x^{24} + x^{36}$.Now,the expression becomes $(1+x^2)^{12}(1 + x^{12} + x^{24} + x^{36})$.Using the binomial theorem,$(1+x^2)^{12} = \sum_{k=0}^{12} {}^{12}C_k (x^2)^k = \sum_{k=0}^{12} {}^{12}C_k x^{2k}$.We need the coefficient of $x^{24}$ in the product $(\sum_{k=0}^{12} {}^{12}C_k x^{2k})(1 + x^{12} + x^{24} + x^{36})$.This is obtained by summing the coefficients of terms that result in $x^{24}$:$1$. From $1 	imes ({}^{12}C_{12} x^{24}) = {}^{12}C_{12} x^{24} = 1 \cdot x^{24}$.$2$. From $x^{12} 	imes ({}^{12}C_6 x^{12}) = {}^{12}C_6 x^{24}$.$3$. From $x^{24} 	imes ({}^{12}C_0 x^0) = {}^{12}C_0 x^{24} = 1 \cdot x^{24}$.Summing these,the coefficient of $x^{24}$ is ${}^{12}C_6 + 1 + 1 = {}^{12}C_6 + 2$.

The coefficient of $x^{24}$ in the expansion of $(1+x^2)^{12}(1+x^{12})(1+x^{24})$ is

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