JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Sequences and Series

The sum of the series

\sum_{n = 1}^{\infty} \frac{n ^{2} + 6 n + 10}{( 2 n + 1 )!}

is equal to :

Answer: (B)

Solution

Put

k = 2 n + 1

, so

n = \frac{k - 1}{2}

and

n^{2} + 6 n + 10 = \frac{( k - 1 ) ^{2}}{4} + 3 (k - 1) + 10 = \frac{k ^{2} + 10 k + 29}{4}

Writing

k^{2} = k (k - 1) + k

splits each term into reciprocals of factorials:

\frac{n ^{2} + 6 n + 10}{( 2 n + 1 )!} = \frac{k ( k - 1 ) + 11 k + 29}{4 k !} = \frac{1}{4} [\frac{1}{( k - 2 )!} + \frac{11}{( k - 1 )!} + \frac{29}{k !}]

n

runs through

1, 2, 3, \dots

, the index

k

runs through the odd integers

3, 5, 7, \dots

, so

k - 2

runs through the odd integers from

1

and

k - 1

through the even integers from

2

. With

\frac{e - e ^{- 1}}{2}

summing the odd reciprocal factorials and

\frac{e + e ^{- 1}}{2}

the even ones,

\sum \frac{1}{( k - 2 )!} = \frac{e - e ^{- 1}}{2}, \sum \frac{1}{( k - 1 )!} = \frac{e + e ^{- 1}}{2} - 1, \sum \frac{1}{k !} = \frac{e - e ^{- 1}}{2} - 1

Collecting the coefficients of

e

and

e^{- 1}

gives

\frac{1}{4} [\frac{41}{2} e - \frac{19}{2} e^{- 1} - 40]

so the series sums to

\frac{41}{8} e - \frac{19}{8} e^{- 1} - 10

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option