1 + frac{2^4}{2!} + frac{3^4}{3!} + frac{4^4} | vedclass

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(C) The general term of the series is $T_n = \frac{n^4}{n!}$ for $n \ge 1$.We know that $\sum_{n=1}^{\infty} \frac{n^4}{n!} = \sum_{n=1}^{\infty} \fra

Vedclass · Accepted Answer

$15e$

Vedclass · Answer

(C) The general term of the series is $T_n = \frac{n^4}{n!}$ for $n \ge 1$.We know that $\sum_{n=1}^{\infty} \frac{n^4}{n!} = \sum_{n=1}^{\infty} \frac{n^3}{(n-1)!}$.Let $m = n-1$,then $n = m+1$. The sum becomes $\sum_{m=0}^{\infty} \frac{(m+1)^3}{m!} = \sum_{m=0}^{\infty} \frac{m^3 + 3m^2 + 3m + 1}{m!}$.Using the identity $\sum_{m=0}^{\infty} \frac{m^k}{m!} = B_k e$,where $B_k$ are Bell numbers:For $k=0$,$\sum \frac{1}{m!} = e$.For $k=1$,$\sum \frac{m}{m!} = e$.For $k=2$,$\sum \frac{m^2}{m!} = 2e$.For $k=3$,$\sum \frac{m^3}{m!} = 5e$.Substituting these values:Sum $= 5e + 3(2e) + 3(e) + e = 5e + 6e + 3e + e = 15e$.

$1 + \frac{2^4}{2!} + \frac{3^4}{3!} + \frac{4^4}{4!} + \dots \infty = $

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