The sum ₙ=1}^{∞} 2 n²+3 n+4/(2n)! is equal to : [1-Feb-2023 Shift 2]

Correct Answer is : 13e4+54e−413e/4 + 5/4e - 4413e+4e5−4

Correct Answer is : 13e4+54e−413e/4 + 5/4e - 4413e+4e5−4

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JEE Mains 1 Feb 2023 Shift 2 Paper

Section: Mathematics

Question : 61 of 90

Marks: +1, -0

The sum

\sum\limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2n)!}

Solution:

\;\sum\limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2n)!}

\;\frac{1}{2} \sum\limits_{n=1}^{\infty} \frac{2n(2n-1)+8n+8}{(2n)!}

\frac{1}{2} \sum\limits_{n=1}^{\infty} \frac{1}{(2n-2)!} + 2 \sum\limits_{n=1}^{\infty} \frac{1}{(2n-1)!} + 4 \sum\limits_{n=1}^{\infty} \frac{1}{(2n)!}

e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots

e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} + \dots

\left(e + \frac{1}{e}\right) = 2 \left(1 + \frac{1}{2!} + \frac{1}{4!} + \dots\right)

e - \frac{1}{e} = \left(1 + \frac{1}{3!} + \frac{1}{5!} + \dots\right)

Now

\frac{1}{2} \left(\sum\limits_{n=1}^{\infty} \frac{1}{(2n-2)!}\right) + 2 \sum\limits_{n=1}^{\infty} \frac{1}{(2n-1)!} + 4 \sum\limits_{n=1}^{\infty} \frac{1}{(2n)!}

= \frac{1}{2} \left[\frac{e+\frac{1}{e}}{2}\right] + 2 \left[\frac{e-\frac{1}{e}}{2}\right] + 4 \left(\frac{e+\frac{1}{e}-2}{2}\right)

= \frac{e+\frac{1}{e}}{4} + e - \frac{1}{e} + 2e + \frac{2}{e} - 4

= \frac{13}{4}e + \frac{5}{4e} - 4

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