Question

If $\frac{5}{1!}+\frac{11}{2!}+\frac{19}{3!}+\frac{29}{4!}+\frac{41}{5!}+\cdots \infty =a{e}^b+c,$
$\text{then}a+b+c=$

Answer 1

$A_n=5+11+19+29+41+\cdots +t_n{A}_n=5+11+19+29+41+\cdots +t_{n-1}+t_n$
On subtracting, we get
$0=5+6+8+10+12+\cdots -t_n$
$t_n=5+[6+8+10+12+\cdots (n-1\left)\text{terms}\right]$
$=5+\left(\frac{n-1}{2}\right)[6\times 2+(n-2)\times 2]$
$t_n=n^2+3n+1$
Therefore, $nth$ term of the given series is,
$T_n=\frac{n^2+3n+1}{n!}$
$=\frac{n^2}{n(n-1)!}+\frac{3n}{n(n-1)!}+\frac{1}{n!}$
$=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}+\frac{3}{(n-1)!}+\frac{1}{n!}$
$=\frac{1}{(n-2)!}+\frac{4}{(n-1)!}+\frac{1}{n!}$
$\text{Now,}{S}_n=\sum T_n$
$S_n=\sum _{n=2}^{\infty }\frac{1}{(n-2)!}+4\sum _{n=1}^{\infty }\frac{1}{(n-1)!}+\sum _{n=1}^{\infty }\frac{1}{n!}$

$\because e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$
$\therefore S_n=e+4e+e-1=6e-1$
$\Rightarrow a=6,b=1,c=-1\therefore a+b+c=6$