Exponential and logarithmic series: Find the sum of $2^2 + 3^2/ 2!+4^2/3!+...$ to infinity 
Find the sum of the following series:
  $ 2^2 + 3^2/2! + 4^2/3! + ...$ to infinity

The answer is given as $5e$ but I got it as $5e+1$
 $T_n = 1/(n-2)! +3/(n-1)! + 1/(n)! $ for $n \ge 2$
and $T_1+T_2 + ... $ to infinity = $4 + e + 3(e-1) +(e-2)$
$=5e-1$
Can you teell me which is correct and how?

The answer is $5e-1$. There might be some confusion with the initial values of the sum.
By letting $T_n = \frac{1}{(n-2)!} +  \frac{3}{(n-1)!} +  \frac{1}{n!}$, as you did:
$S = 2^2 + T_2 + T_3 + \dots = 4 + (\frac{1}{0!} +  \frac{3}{1!} +  \frac{1}{2!}) + (\frac{1}{1!} +  \frac{3}{2!} +  \frac{1}{3!}) + (\frac{1}{2!} +  \frac{3}{3!} +  \frac{1}{4!}) + \dots$
Then we can take the $\frac{1}{0!}$, $\frac{3}{1!}$, and $\frac{1}{1!}$ out of the brackets, giving us $S=4 + \frac{1}{0!} + \frac{3}{1!} + \frac{1}{1!} + 5(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots) = 4+1+3+1+5(e-2) = 5e-1$.
