Question

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

• A. $S<8$
• B. $S>12$
• C. $8<S<12$
• D. $S=8$

Answer 1

Answer: (C)

$8<S<12$

Let $S=1+\frac{8}{2!}+\frac{21}{3!}+\frac{40}{4!}+\frac{65}{5!}+\dots$
Again,

let $S_1=1+8+21+40+65+\cdots +T_n$

$S_1=1+8+21+40+\cdots +T_n$

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$0=1+7+13+19+25+\cdots -T_n$

$T_n=1+7+13+19+25+\cdots +n$

$=\frac{n}{2}[2\left(1\right)+(n-1)6]$

$=n[1+3(n-1)]=n(3n-2)$

$\therefore S=\sum \frac{n(3n-2)}{n!}$

$=\sum \frac{3n-2}{(n-1)!}$

$=\sum \frac{3n-3+1}{(n-1)!}$

$S=\sum \frac{3}{(n-2)!}+\sum \frac{1}{(n-1)!}$

$=3e+e$ $[\because e=1+\frac{1}{1!}+\frac{1}{2!}+\dots ]$

$=4e$

We know $2<e<3$
$\therefore 8<4e<12$
$\Rightarrow 8<S<12$