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JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Sequences and Series

Let

S_{n} = 1 \cdot (n - 1) + 2 \cdot (n - 2) + 3 \cdot (n - 3) + \dots + (n - 1) \cdot 1, n \geq 4

. The sum

\sum_{n = 4}^{\infty} (\frac{2 S _{n}}{n !} - \frac{1}{( n - 2 )!})

is equal to:

(A) $\frac{e - 2}{6}$
(B) $\frac{e}{3}$
(C) $\frac{e - 1}{3}$ ✓ Correct
(D) $\frac{e}{6}$

Answer: (C)

Solution

Here

S_{n} = \sum_{k = 1}^{n - 1} k (n - k) = \frac{n ( n - 1 ) ( n + 1 )}{6}

, so

\frac{2 S _{n}}{n !} = \frac{n ( n - 1 ) ( n + 1 )}{3 n !} = \frac{n + 1}{3 ( n - 2 )!} .

Then

\frac{2 S _{n}}{n !} - \frac{1}{( n - 2 )!} = \frac{( n + 1 ) - 3}{3 ( n - 2 )!} = \frac{n - 2}{3 ( n - 2 )!} = \frac{1}{3 ( n - 3 )!} .

Summing from

n = 4

with

m = n - 3

,

n = 4 \sum \infty \frac{1}{3 ( n - 3 )!} = \frac{1}{3} m = 1 \sum \infty \frac{1}{m !} = \frac{e - 1}{3} .

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option

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