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

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Question

(A) The expression is $(1 + t^2)^{12}(1 + t^{12})(1 + t^{24})$.Expanding $(1 + t^2)^{12}$ using the binomial theorem,we get $\sum_{k=0}^{12} {^{12}C_k

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$^{12}C_6 + 2$

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(A) The expression is $(1 + t^2)^{12}(1 + t^{12})(1 + t^{24})$.Expanding $(1 + t^2)^{12}$ using the binomial theorem,we get $\sum_{k=0}^{12} {^{12}C_k} (t^2)^k = \sum_{k=0}^{12} {^{12}C_k} t^{2k}$.We want the coefficient of $t^{24}$ in the product $(\sum_{k=0}^{12} {^{12}C_k} t^{2k})(1 + t^{12} + t^{24})$.Distributing the terms:$1$. From $1 	imes (\dots)$,we need the coefficient of $t^{24}$ in $(1 + t^2)^{12}$,which is $^{12}C_{12} = 1$.$2$. From $t^{12} 	imes (\dots)$,we need the coefficient of $t^{12}$ in $(1 + t^2)^{12}$,which is $^{12}C_6 = 924$.$3$. From $t^{24} 	imes (\dots)$,we need the coefficient of $t^0$ in $(1 + t^2)^{12}$,which is $^{12}C_0 = 1$.Summing these coefficients: $1 + ^{12}C_6 + 1 = ^{12}C_6 + 2$.

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

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