Question

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

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

Answer 1

Answer: (C)

Let $S=1+\frac{8}{2!}+\frac{21}{3!}+\frac{40}{4!}+\frac{65}{5!}+\dots$
Again, let $S_1=1+8+21+40+65+\dots +T_n$
and $\frac{S_1=+1+8+21+40+\dots +T_n}{0=1+7+13+19+25+\dots -T_n}$
$T_n=1+7+13+19+25+\dots +n$ terms
$=\frac{n}{2}[2(1)+(n-1)6]$
$=n[1+3(n-1)]=n(3n-2)$
$\therefore S=\sum \frac{n(3n-2)}{n!}$
$=\sum \frac{3n-2}{(n-1)!}$
$=\sum \frac{3n-3+1}{(n-1)!}$
$S=\sum \frac{3}{(n-2)!}+\sum \frac{1}{(n-1)!}$
$=3e+e\left[\because e=1+\frac{1}{1!}+\frac{1}{2!}+\dots \right]$
$=4e$
We know $2<e<3$
$\therefore 8<4e<12$
$\Rightarrow 8<S<12$