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$1+\frac{1+2}{2 !}+\frac{1+2+2^2}{3 !}+\ldots$ is equal to
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2909 Upvotes
Verified Answer
The correct answer is:
$e^2-e$
We have,
$\begin{gathered}
1+\frac{1+2}{2 !}+\frac{1+2+2^2}{3 !}+\ldots \\
T_n=\frac{1+2+2^2+\ldots+2^{n-1}}{n !} \\
=\frac{\frac{1\left(2^n-1\right)}{2-1}}{n !}=\frac{2^n-1}{n !} \\
T_n=\frac{2^n}{n !}-\frac{1}{n !} \\
S_n=\Sigma T_n=\Sigma \frac{2^n}{n !}-\frac{1}{n !}
\end{gathered}$
$\begin{gathered}=\left(\frac{1}{1}+\frac{2}{1 !}+\frac{2^2}{2 !}+\ldots\right)-\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right) \\ =e^2-e\end{gathered}$
$\begin{gathered}
1+\frac{1+2}{2 !}+\frac{1+2+2^2}{3 !}+\ldots \\
T_n=\frac{1+2+2^2+\ldots+2^{n-1}}{n !} \\
=\frac{\frac{1\left(2^n-1\right)}{2-1}}{n !}=\frac{2^n-1}{n !} \\
T_n=\frac{2^n}{n !}-\frac{1}{n !} \\
S_n=\Sigma T_n=\Sigma \frac{2^n}{n !}-\frac{1}{n !}
\end{gathered}$
$\begin{gathered}=\left(\frac{1}{1}+\frac{2}{1 !}+\frac{2^2}{2 !}+\ldots\right)-\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right) \\ =e^2-e\end{gathered}$
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