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

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Question

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

Vedclass · Accepted Answer

$1 + ^{25}C_5$

Vedclass · Answer

(A) We need the coefficient of $t^{50}$ in the expansion of $(1 + t^2)^{25}(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$.Expanding $(1 + t^2)^{25} = \sum_{k=0}^{25} {^{25}C_k} (t^2)^k = \sum_{k=0}^{25} {^{25}C_k} t^{2k}$.To obtain $t^{50}$,we consider the product of terms from $(1 + t^2)^{25}$ and the second factor $(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$:$1$. Multiplying $1$ from the second factor with the term $t^{50}$ from $(1 + t^2)^{25}$ (where $k=25$): Coefficient = $^{25}C_{25} = 1$.$2$. Multiplying $t^{40}$ from the second factor with the term $t^{10}$ from $(1 + t^2)^{25}$ (where $k=5$): Coefficient = $^{25}C_5$.Other combinations like $t^{25} 	imes t^{25}$ are not possible because $(1 + t^2)^{25}$ only contains even powers of $t$.Thus,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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