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  4. Let S denote the sum of the infinite series 1+(8/2!)+(21/3!)+(40/4!)+(65/5!) ....... . Then

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Q. Let $S$ denote the sum of the infinite series $1+\frac{8}{2!}+\frac{21}{3!}+\frac{40}{4!}+\frac{65}{5!} ....... $. Then

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Solution:

Let $S=1+\frac{8}{2 !}+\frac{21}{3 !}+\frac{40}{4 !}+\frac{65}{5 !}+\ldots$
Again, let $S_{1}=1+8+21+40+65+\ldots+T_{n}$
and $\frac{S_{1}=+1+8+21+40+\ldots+T_{n}}{0=1+7+13+19+25+\ldots-T_{n}}$
$T_{n}=1+7+13+19+25+\ldots+n$ terms
$=\frac{n}{2}[2(1)+(n-1) 6]$
$=n[1+3(n-1)]=n(3 n-2)$
$\therefore S=\sum \frac{n(3 n-2)}{n !}$
$=\sum \frac{3 n-2}{(n-1) !}$
$=\sum \frac{3 n-3+1}{(n-1) !}$
$S=\sum \frac{3}{(n-2) !}+\sum \frac{1}{(n-1) !}$
$=3 e+e \,\,\,\,\left[\because e=1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right]$
$=4 e$
We know $ 2 < e < 3 $
$ \therefore 8 < 4 e < 12 $
$ \Rightarrow 8 < S < 12$