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The sum  \sum_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !} is equal to :

Option: 1

\frac{13 \mathrm{e}}{4}+\frac{5}{4 \mathrm{e}}


Option: 2

\frac{11 \mathrm{e}}{2}+\frac{7}{2 \mathrm{e}}-4


Option: 3

\frac{11 \mathrm{e}}{2}+\frac{7}{2 \mathrm{e}}


Option: 4

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


Answers (1)

best_answer

\begin{aligned} & \sum_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !} \\ & \frac{1}{2} \sum_{n=1}^{\infty} \frac{2 n(2 n-1)+8 n+8}{(2 n) !} \\ & \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}+2 \sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \sum_{n=1}^{\infty} \frac{1}{(2 n) !} \end{aligned}

\begin{aligned} & \mathrm{e}=1+1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots \\ & e^{-1}=1-1+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots \\ & \left(e+\frac{1}{e}\right)=2\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots \ldots\right) \end{aligned}

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

\begin{aligned} & \frac{1}{2}\left(\sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}\right)+2 \sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \sum_{n=1}^{\infty} \frac{1}{(2 n) !} \\ & =\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) \end{aligned}

\begin{aligned} & =\frac{\left(e+\frac{1}{e}\right)}{4}+e-\frac{1}{e}+2 e+\frac{2}{e}-4 \\ & =\frac{13}{4} e+\frac{5}{4 e}-4 \end{aligned}

Posted by

Sanket Gandhi

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