Evaluate $\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......$ 
Prove that $$\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......=12e-5$$

$$
e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+......
$$
I have no clue of where to start and I am not able to find the general expression of the nth term.

Look at the sequence of numerators and their differences
\begin{array}
n &1 & 2 &3 &4& 5 \\
u_n & 9 & 19 &35&57&85 \\
1st diff. & 10&16&22&28& \\
2nd diff. &6&6&6&&
\end{array}
The first differences show the sequence is not of the form $an+b$, i.e., not a linear sequence. Working out the $2$nd differences gives all the same difference of $6$ and this means our sequence is of the form $an^2+bn+c$, i.e., a quadratic sequence.
Now $a=1/2 * 2$nd difference, which for us is $a = 3$.
Our sequence is then of the form: $3n^2+bn+c$. 
Plug in $n=1$ to get $3*1^2+b*1+c = 3+b+c=9$, so $b+c=6$. 
Plug in $n=2$ to get $3*2^2+b*2+c = 12+2b+c = 19$, so $2b+c=7$. Hence $b=1$, $c=5$. So our sequence is $3n^2+n+5$.
Set this series equal to the first and rearrange
\begin{align}
\sum_{n=1}^{\infty} \frac{3n^2+n+5}{n!}&=3\sum_{n=1}^{\infty} \frac{n}{(n-1)!}+\sum_{n=1}^{\infty} \frac{1}{(n-1)!}+5\sum_{n=1}^{\infty} \frac{1}{n!}\\
&=12+12\sum_{n=1}^{\infty} \frac{1}{n!}-5=7+12\sum_{n=1}^{\infty} \frac{1}{n!}
\end{align}
Rearrange to
$$3\sum_{n=1}^{\infty} \frac{n}{(n-1)!}+\sum_{n=1}^{\infty} \frac{1}{(n-1)!}=7+7\sum_{n=1}^{\infty} \frac{1}{n!}=7e$$
now subtract
$$3\sum_{n=1}^{\infty} \frac{n}{(n-1)!}=7e-\sum_{n=0}^{\infty} \frac{1}{n!}=6e$$
and divide
\begin{align}
\sum_{n=1}^{\infty} \frac{n}{(n-1)!}=\sum_{n=0}^{\infty} \frac{1+n}{n!}=\sum_{n=0}^{\infty} \frac{1}{n!}+\sum_{n=0}^{\infty} \frac{n}{n!}=\sum_{n=0}^{\infty} \frac{1}{n!}+\sum_{n=1}^{\infty} \frac{1}{(n-1)!}=2e
\end{align}
and everything matches up as required.
