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Let $S$ denote the sum of the infinite series $1+\frac{8}{2 !}+\frac{21}{3 !}+\frac{40}{4 !}+\frac{65}{5 !}+\ldots .$ Then
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The correct answer is:
$8 < S < 12$
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 $\quad S_{1}=+1+8+21+40+\ldots+T_{n}$
$$
\begin{aligned} 0=1+7+13+19+25+\ldots-T_{n} \\ T_{n} &=1+7+13+19+25+\ldots+n \text { terms } \\ &=\frac{n}{2}[2(1)+(n-1) 6] \\ &=n[1+3(n-1)]=n(3 n-2) \end{aligned}
$$
$S=\sum \frac{3 n-2}{(n-1) !}$
$=\sum \frac{3 n-3+1}{(n-1) !}$
$=\sum \frac{3}{(n-2) !}+\sum \frac{1}{(n-1) !}$
$\quad=3 e+e \quad\left[\because e=1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right]$
$=4 e$
We know $\quad 2 < e < 3$
$\therefore \quad 8 < 4 e < 12$
$\Rightarrow \quad 8 < S < 12$
Again, let $S_{1}=1+8+21+40+65+\ldots+T_{n}$
and $\quad S_{1}=+1+8+21+40+\ldots+T_{n}$
$$
\begin{aligned} 0=1+7+13+19+25+\ldots-T_{n} \\ T_{n} &=1+7+13+19+25+\ldots+n \text { terms } \\ &=\frac{n}{2}[2(1)+(n-1) 6] \\ &=n[1+3(n-1)]=n(3 n-2) \end{aligned}
$$
$S=\sum \frac{3 n-2}{(n-1) !}$
$=\sum \frac{3 n-3+1}{(n-1) !}$
$=\sum \frac{3}{(n-2) !}+\sum \frac{1}{(n-1) !}$
$\quad=3 e+e \quad\left[\because e=1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right]$
$=4 e$
We know $\quad 2 < e < 3$
$\therefore \quad 8 < 4 e < 12$
$\Rightarrow \quad 8 < S < 12$
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