Q52 / 160 · TS EAMCET 2008 · Mathematics
$\sum_{k=1}^{\infty} \frac{1}{k !}\left(\sum_{n=1}^k 2^{n-1}\right)$ is equal to
- A$e$
- B$e^2+e$
- C$e^2$
- Correct answerD$e^2-e$
Solution
$\begin{aligned} & \sum_{k=1}^{\infty} \frac{1}{k !}\left(\sum_{n=1}^k 2^{n-1}\right) \\ &=\sum_{k=1}^{\infty} \frac{1}{k !}\left[1\left(2^k-1\right)\right]\end{aligned}$
$\begin{aligned} & =\sum_{k=1}^{\infty} \frac{2^k-1}{k !} \\ & =\sum_{k=1}^{\infty} \frac{2^k}{k !}-\sum_{k=1}^{\infty} \frac{1}{k !} \\ & =e^2-1-(e-1) \\ & =e^2-e\end{aligned}$
Hence, the correct answer is option: 'D'.