AP EAMCET · Maths · Sequences and Series
\(\sum_{k=1}^{\infty} \frac{1}{k !}\left(\sum_{n=1}^k 2^{n-1}\right)\) is equal to
- A \(e\)
- B \(e^2+e\)
- C \(e^2\)
- D \(e^2-e\)
Answer & Solution
Correct Answer
(D) \(e^2-e\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sum_{k=1}^{\infty} \frac{1}{k !}\left(\sum_{n=1}^k 2^{n-1}\right) \\ &=\sum_{k=1}^{\infty} \frac{1}{k !}\left[1\left(2^k-1\right)\right]\end{aligned}\)…
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