∑ _ n = 0 ^ ∞ n ^ 3 ( ( 2n ) ! ) + ( 2n - 1 ) ( n ! ) ( n ! ) ( ( 2n ) ! ) = ae + (b)/(e) + c where a, b, c ∈ ℤ and e = ∑ _ n = 0 ^ ∞ (1)/(n !) Then a ^ 2 - b + c is equal to _______

Correct answer: 0 Given: ∑ _ n = 0 ^ ∞ [ n ^ 3 \ (2n) !\ + (2n - 1)n ! n ! × (2n) ! ] = ∑ _ n = 0 ^ ∞ [ n ^ 3 \ (2n) !\ n ! × (2n) ! ] + ∑ _ n = 0 ^ ∞ [((2n - 1)n !)/(n ! × (2n) !)] = ∑ _ n = 0 ^ ∞ [ n ^ 3 n ! ] + ∑ _ n = 0 ^ ∞ [((2n - 1))/((2n) !)] = ∑ _ n = 0 ^ ∞ [(n(n - 1)(n - 2) + 3n(n - 1) + n)/(n !)] + ∑ _ n = 1 ^ ∞ [(1)/((2n - 1) !)] - ∑ _ n = 0 ^ ∞ [(1)/((2n) !)] = ∑ _ n = 0 ^ ∞ [(n(n - 1)(n - 2))/(n !)] + ∑ _ n = 0 ^ ∞ [(3n(n - 1))/(n !)] + ∑ _ n = 0 ^ ∞ [(n)/(n !)] + ( e - e ^ - 1 2 ) - ( e + e ^ - 1 2 ) = ∑ _ n = 3 ^ ∞ [(1)/((n - 3) !)] + 3 ∑ _ n = 2 ^ ∞ [(1)/((n - 2) !)] + ∑ _ n = 1 ^ ∞ [(1)/((n - 1) !)] - (1)/(e) = 5e - (1)/(e) So, a = 5, b = - 1, c = 0 Hence, a ^ 2 - b + c = 25 + 1 = 26

JEE Main 202330 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 30 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2023 · 30 January · Morning Shift · Mathematics · Sequences and Series

∑ _ n = 0 ^ ∞ n ^ 3 ( ( 2n ) ! ) + ( 2n - 1 ) ( n ! ) ( n ! ) ( (...

\mathop{∑}_{n = 0}^{\infty}\dfrac{{n}^{3} ( ( 2n ) ! ) + ( 2n - 1 ) ( n ! )}{( n ! ) ( ( 2n ) ! )} = ae + \dfrac{b}{e} + c

where

a, b, c \in \mathbb{Z}

and

e = \mathop{∑}_{n = 0}^{\infty}\dfrac{1}{n !}

Then

{a}^{2} - b + c

is equal to _______

Correct answer

Step-by-step solution

Given:

$\mathop{∑}_{n = 0}^{\infty}\left[\dfrac{{n}^{3}\left\{\left(2n\right) !\right\} + \left(2n - 1\right)n !}{n ! \times \left(2n\right) !}\right]$

$= \mathop{∑}_{n = 0}^{\infty}\left[\dfrac{{n}^{3}\left\{\left(2n\right) !\right\}}{n ! \times \left(2n\right) !}\right] + \mathop{∑}_{n = 0}^{\infty}\left[\dfrac{\left(2n - 1\right)n !}{n ! \times \left(2n\right) !}\right]$

$= \mathop{∑}_{n = 0}^{\infty}\left[\dfrac{{n}^{3}}{n !}\right] + \mathop{∑}_{n = 0}^{\infty}\left[\dfrac{\left(2n - 1\right)}{\left(2n\right) !}\right]$

$= \mathop{∑}_{n = 0}^{\infty}\left[\dfrac{n\left(n - 1\right)\left(n - 2\right) + 3n\left(n - 1\right) + n}{n !}\right] + \mathop{∑}_{n = 1}^{\infty}\left[\dfrac{1}{\left(2n - 1\right) !}\right] - \mathop{∑}_{n = 0}^{\infty}\left[\dfrac{1}{\left(2n\right) !}\right]$

$= \mathop{∑}_{n = 0}^{\infty}\left[\dfrac{n\left(n - 1\right)\left(n - 2\right)}{n !}\right] + \mathop{∑}_{n = 0}^{\infty}\left[\dfrac{3n\left(n - 1\right)}{n !}\right] + \mathop{∑}_{n = 0}^{\infty}\left[\dfrac{n}{n !}\right] + \left(\dfrac{e - {e}^{- 1}}{2}\right) - \left(\dfrac{e + {e}^{- 1}}{2}\right)$

$= \mathop{∑}_{n = 3}^{\infty}\left[\dfrac{1}{\left(n - 3\right) !}\right] + 3\mathop{∑}_{n = 2}^{\infty}\left[\dfrac{1}{\left(n - 2\right) !}\right] + \mathop{∑}_{n = 1}^{\infty}\left[\dfrac{1}{\left(n - 1\right) !}\right] - \dfrac{1}{e}$

$= 5e - \dfrac{1}{e}$

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

Hence, ${a}^{2} - b + c = 25 + 1 = 26$

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