TS EAMCET · Maths · Sequences and Series
\(\sum_{n=1}^{\infty} \frac{2 n^2+n+1}{n !}\) is equal to
- A \(2 e-1\)
- B \(2 e+1\)
- C \(6 e-1\)
- D \(6 e+1\)
Answer & Solution
Correct Answer
(C) \(6 e-1\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Let } S=\sum_{n=1}^{-} \frac{2 n^2+n+1}{n !} \\ & =\bar{\sum}_{n=1}\left(\frac{2 n}{(n-1) !}+\frac{1}{(n-1) !}+\frac{1}{n !}\right) \\ & =\bar{\sum}_{n=1}\left(\frac{2}{(n-2) !}+\frac{3}{(n-1) !}+\frac{1}{n !}\right) \\ & =2\left(1+\frac{1}{1…
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