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Sum of (2^k - 1)/k! to Infinity | JEE

JEE Maths question with a full step-by-step solution.

Question
The value of the series is ∑k=1∞1k!(∑n=1k2n−1)\displaystyle\sum_{k=1}^{\infty}\frac{1}{k!}\left(\sum_{n=1}^{k}2^{n-1}\right)
Aee
Be2+ee^2+e
Ce2e^2
De2−ee^2-ecorrect
Solution
Step 1: The inner sum is 1+2+22+⋯+2k−1=2k−1.1+2+2^2+\cdots+2^{k-1}=2^k-1. Step 2: So
∑k=1∞2k−1k!=∑k=1∞2kk!−∑k=1∞1k!=(e2−1)−(e−1)=e2−e.\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.
 Correct answer: (4)
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