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TS EAMCET · Maths · Sequences and Series

\(\sum_{k=1}^{\infty} \frac{1}{k !}\left(\sum_{n=1}^k 2^{n-1}\right)\) is equal to

  1. A \(e\)
  2. B \(e^2+e\)
  3. C \(e^2\)
  4. D \(e^2-e\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(e^2-e\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \sum_{k=1}^{\infty} \frac{1}{k !}\left(\sum_{n=1}^k 2^{n-1}\right) \\ &=\sum_{k=1}^{\infty} \frac{1}{k !}\left[1\left(2^k-1\right)\right]\end{aligned}\)…