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$\frac{2}{2 !}+\frac{2+4}{3 !}+\frac{2+4+6}{4 !}+\ldots$ is equal to
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Verified Answer
The correct answer is:
$e$
We have,
$\begin{aligned} T_n & =\frac{2+4+6+\ldots+2 n}{(n+1) !} \\ T_n & =\frac{\sum 2 n}{(n+1) !}=\frac{2 \frac{n(n+1)}{3}}{(n+1) n(n-1) !} \\ & =\frac{1}{(n-1) !}\end{aligned}$
$\begin{aligned} \therefore \quad S & =T_1+T_2+T_3+\ldots \\ & =1+1+\frac{1}{2 !}+\frac{1}{3 !}+\ldots=e\end{aligned}$
$\begin{aligned} T_n & =\frac{2+4+6+\ldots+2 n}{(n+1) !} \\ T_n & =\frac{\sum 2 n}{(n+1) !}=\frac{2 \frac{n(n+1)}{3}}{(n+1) n(n-1) !} \\ & =\frac{1}{(n-1) !}\end{aligned}$
$\begin{aligned} \therefore \quad S & =T_1+T_2+T_3+\ldots \\ & =1+1+\frac{1}{2 !}+\frac{1}{3 !}+\ldots=e\end{aligned}$
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