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

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Question

(A) Let the expression be $P(t) = (1 + t^2)^{25} (1 + t^{25} + t^{40} + t^{45} + t^{47})$.
We need the coefficient of $t^{50}$ in the expansion of $P

Vedclass · Accepted Answer

$1 + ^{25}C_5$

Vedclass · Answer

(A) Let the expression be $P(t) = (1 + t^2)^{25} (1 + t^{25} + t^{40} + t^{45} + t^{47})$.
We need the coefficient of $t^{50}$ in the expansion of $P(t)$.
Expanding $(1 + t^2)^{25} = \sum_{k=0}^{25} {^{25}C_k} (t^2)^k = \sum_{k=0}^{25} {^{25}C_k} t^{2k}$.
We multiply this by $(1 + t^{25} + t^{40} + t^{45} + t^{47})$.
To get $t^{50}$,we consider the following combinations:
$1$. $t^{50}$ from $(1 + t^2)^{25}$ multiplied by $1$: This corresponds to $k=25$,giving $^{25}C_{25} = 1$.
$2$. $t^{25}$ from the second bracket multiplied by $t^{25}$ from $(1 + t^2)^{25}$: This is impossible as $2k=25$ has no integer solution.
$3$. $t^{40}$ from the second bracket multiplied by $t^{10}$ from $(1 + t^2)^{25}$: This corresponds to $2k=10 \implies k=5$,giving $^{25}C_5$.
$4$. $t^{45}$ from the second bracket multiplied by $t^5$ from $(1 + t^2)^{25}$: Impossible.
$5$. $t^{47}$ from the second bracket multiplied by $t^3$ from $(1 + t^2)^{25}$: Impossible.
Thus,the coefficient of $t^{50}$ is $^{25}C_{25} + ^{25}C_5 = 1 + ^{25}C_5$.

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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