The coefficient of t^{50} in (1 + t^2)^{25}(1 | vedclass

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(A) We need the coefficient of $t^{50}$ in the expansion of $(1 + t^2)^{25}(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$.Since $(1 + t^2)^{25} = \s

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$1 + {}^{25}C_5$

Vedclass · Answer

(A) We need the coefficient of $t^{50}$ in the expansion of $(1 + t^2)^{25}(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$.Since $(1 + t^2)^{25} = \sum_{k=0}^{25} {}^{25}C_k (t^2)^k = \sum_{k=0}^{25} {}^{25}C_k t^{2k}$,the expansion only contains even powers of $t$.Thus,we only consider terms from the second part that are even powers of $t$ and less than or equal to $50$.The relevant terms are $1$ and $t^{40}$.For $1$,we need the coefficient of $t^{50}$ in $(1 + t^2)^{25}$,which is ${}^{25}C_{25} = 1$.For $t^{40}$,we need the coefficient of $t^{10}$ in $(1 + t^2)^{25}$,which is ${}^{25}C_5$.Therefore,the total coefficient of $t^{50}$ is $1 + {}^{25}C_5$.

The coefficient of $t^{50}$ in $(1 + t^2)^{25}(1 + t^{25})(1 + t^{40})(1 + t^{45})(1 + t^{47})$ is

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