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JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Sequences and Series

Let

\sum_{n = 0}^{\infty} \frac{n ^{3} (( 2 n )!) + ( 2 n - 1 ) ( n !)}{( n !) (( 2 n )!)} = a e + \frac{b}{e} + c

,
where

a, b, c \in Z

and

e = \sum_{n = 0}^{\infty} \frac{1}{n !}

. Then

a^{2} - b + c

is equal to ____.

Answer: 26

Solution

Splitting the term,

\frac{n ^{3} ( 2 n )! + ( 2 n - 1 ) ( n !)}{( n !) ( 2 n )!} = \frac{n ^{3}}{n !} + \frac{2 n - 1}{( 2 n )!}

. The first series sums to

\sum_{n \geq 0} \frac{n ^{3}}{n !} = 5 e

. For the second,

\sum_{n \geq 0} \frac{2 n}{( 2 n )!} = \sum_{n \geq 1} \frac{1}{( 2 n - 1 )!} = sinh 1

and

\sum_{n \geq 0} \frac{1}{( 2 n )!} = cosh 1

, so

\sum_{n \geq 0} \frac{2 n - 1}{( 2 n )!} = sinh 1 - cosh 1 = - \frac{1}{e}

. Hence the total is

5 e - \frac{1}{e}

, giving

a = 5, b = - 1, c = 0

and

a^{2} - b + c = 25 + 1 + 0 = 26.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer

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