Question

The value of
$\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+...\infty$ is $xe-y$ Find $x-y$

Answer 1

$\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+...\infty$
Let $S=\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+...\infty$
and $S_1=9+19+35+57+85+.....t_n$ ....(1)
$S_1=9+19+35+57+85+....+t_n$ ...(2)
Subtracting (2) from (1), we get
$0=9+10+16+22+28+....+t_n$
or $t_n=9+[10+16+22+28+....upto(n-1\left)terms\right]$
$\Rightarrow t_n=9+\frac{(n-1)}{2}(20+(n-2\left)6\right)$
$\Rightarrow t_n=9+(n-1)(3n+4)$
So, $n^{th}$ terms of the given series is
$T_n=\frac{9+(n-1)(3n+4)}{n!}$
$\Rightarrow T_n=\frac{3n^2+n+5}{n!}$
$\Rightarrow T_n=\frac{3n^2}{n!}+\frac{n}{n!}+\frac{5}{n!}$
$\Rightarrow T_n=\frac{3n}{(n-1)!}+\frac{1}{(n-1)!}+\frac{5}{n!}$
$\Rightarrow T_n=\frac{3n-3}{(n-1)!}+\frac{3}{(n-1)!}+\frac{1}{(n-1)!}+\frac{5}{n!}$

$\Rightarrow T_n=\frac{3}{(n-2)!}+\frac{4}{(n-1)!}+\frac{5}{n!}$
$S=\sum T_n$
$=3\sum \frac{1}{(n-2)!}+4\sum \frac{1}{(n-1)!}+5\sum \frac{1}{n!}$
$=3e+4e+5(e-1)=12e-5$
Comparing with given value
$x=12,y=5$
$\Rightarrow x-y=7$