Let sum_{n=0}^{infty} frac{n^3((2n)!) + (2n-1 | vedclass

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(D) The given expression is $\sum_{n=0}^{\infty} \left( \frac{n^3}{(n!)} + \frac{2n-1}{(2n)!} \right)$.We know that $n^3 = n(n-1)(n-2) + 3n(n-1) + n$.

Vedclass · Accepted Answer

$26$

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(D) The given expression is $\sum_{n=0}^{\infty} \left( \frac{n^3}{(n!)} + \frac{2n-1}{(2n)!} \right)$.We know that $n^3 = n(n-1)(n-2) + 3n(n-1) + n$.Thus,$\sum_{n=0}^{\infty} \frac{n^3}{n!} = \sum_{n=3}^{\infty} \frac{1}{(n-3)!} + 3\sum_{n=2}^{\infty} \frac{1}{(n-2)!} + \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = e + 3e + e = 5e$.For the second part,$\sum_{n=0}^{\infty} \frac{2n-1}{(2n)!} = \sum_{n=1}^{\infty} \frac{2n}{(2n)!} - \sum_{n=0}^{\infty} \frac{1}{(2n)!} = \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} - \sum_{n=0}^{\infty} \frac{1}{(2n)!}$.Using the series for $e$ and $e^{-1}$,$\sum_{n=0}^{\infty} \frac{1}{(2n)!} = \frac{e + e^{-1}}{2}$ and $\sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \frac{e - e^{-1}}{2}$.Note that $\sum_{n=1}^{\infty} \frac{1}{(2n-1)!} = \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} = \frac{e - e^{-1}}{2}$.So,the sum is $5e + \frac{e - e^{-1}}{2} - \frac{e + e^{-1}}{2} = 5e - e^{-1}$.Comparing with $ae + be^{-1} + c$,we get $a=5, b=-1, c=0$.Therefore,$a^2 - b + c = 5^2 - (-1) + 0 = 25 + 1 = 26$.

Let $\sum_{n=0}^{\infty} \frac{n^3((2n)!) + (2n-1)(n!)}{(n!)((2n)!)} = ae + \frac{b}{e} + c$,where $a, b, c \in \mathbb{Z}$ and $e = \sum_{n=0}^{\infty} \frac{1}{n!}$. Then $a^2 - b + c$ is equal to $................$.

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