Question

$\frac{4}{1!}+\frac{11}{2!}+\frac{22}{3!}+\frac{37}{4!}+\frac{56}{5!}+...\infty =be-1$ Find b

Answer 1

Consider the series $4+11+22+37+56+...$
Let $t_n$ be its nth term and $S_n$ be the sum of its first n terms, then
$S_n=4+11+22+37+56+...+t_{n-1}+t_n$ ...(1)
also $S_n=4+11+22+37+...+t_{n-1}+t_n$ ...(2)
writing one term shifted to right.
substracting (2) from (1), then we get
$0=4+7+11+15+19+...-t_n$
$t_n=4+7+11+15+19+...tonterms$
$=4+[7+11+15+19+...to(n-1\left)terms\right]$
$=4+\frac{(n-1)}{2}[2.7+(n-1-1\left)4\right]$
$=4+\frac{(n-1)}{2}[4n-6]$
$=4+(n-1)(2n+3)$
$=2n^2+n+1$
If $T_n$ be the nth term of the given series, then
$T_n=\frac{t_n}{n!}$
$=\frac{2n^2+n+1}{n!}$
$=\frac{2n^2}{n!}+\frac{n}{n!}+\frac{1}{n!}$
$=\frac{2n^2}{n(n-1)!}+\frac{n}{n(n-1)!}+\frac{1}{n!}$
$=\frac{2n}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}$
$=\frac{2\left[\right(n-1)+1]}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}$
$=2(\frac{(n-1)}{(n-1)!}+\frac{1}{(n-1)!})+\frac{1}{(n-1)!}+\frac{1}{n!}$
$=\frac{2}{(n-2)!}+\frac{3}{(n-1)!}+\frac{1}{n!}$
$=\frac{2(n-1)}{(n-1)(n-2)!}+\frac{2}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}$
$=\frac{2}{(n-2)!}+\frac{3}{(n-1)!}+\frac{1}{n!}$
$\therefore$ Sum of the given series.
$S=\sum _{n=1}^{\infty }{T}_n$
$=\frac{2}{(n-2)!}+\frac{3}{(n-1)!}+\frac{1}{n!}$

$=2\sum _{n-1}^{\infty }\frac{2}{(n-2)!}+3\sum _{n-1}^{\infty }\frac{3}{(n-1)!}+\sum _{n-1}^{\infty }\frac{1}{n!}$
$=2\left(e\right)+3\left(e\right)+e-1=6e-1$