JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(S_{n}=1 \cdot(n-1)+2 \cdot(n-2)+3 \cdot(n-3)+\ldots+\) \((\mathrm{n}-1) \cdot 1, \mathrm{n} \geq 4\) The sum \(\sum_{n=4}^{\infty}\left(\frac{2 S_{n}}{n !}-\frac{1}{(n-2) !}\right)\) is equal to :
- A \(\frac{\mathrm{e}-1}{3}\)
- B \(\frac{e-2}{6}\)
- C \(\frac{\mathrm{e}}{3}\)
- D \(\frac{\mathrm{e}}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{e}-1}{3}\)
Step-by-step Solution
Detailed explanation
Let \(T_{r}=r(n-r)\) \(T_{r}=n r-r^{2}\) \(\Rightarrow S_{n}=\sum_{r=1}^{n} T_{r}=\sum_{r=1}^{n}\left(n r-r^{2}\right)\) \(S_{n}=\frac{n \cdot(n)(n+1)}{2}-\frac{n(n+1)(2 n+1)}{6}\) \(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}+1)}{6}\) Now…
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