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.

You noticed a pattern about the denominators, they are straight forward. How do you figure out what's going on with the numerators? 
It's good to check ratios between consecutive terms if they are increasing, but that doesn't get you anywhere here. Anther good rule of thumb, take consecutive differences, perhaps differences in those differences, and so on. If your iterated sequences of differences ends up with 0s, then you know you are dealing with a polynomial of degree $n-1$ where $n$ is the number of differences you needed to take. 
So we have 9,19, 35, 57, 85. 
First layer of differences is 10, 16, 22, 28.
Second Layer: 6,6,6. 
Third Layer: 0,0. 
So we needed three layers to get a layer of zeros, so our numerators are formed by a quadratic function in n. 
With some algebra, you can determined that given a polynomial of $f(x)=ax^2+bx+c$, layers of $f(x+1)-f(x)$, the second layer will be a constant,term by term, and specifically $2a$. Six is our iterated constant and $2a=6$, so the coefficient of $n^2$ must be 3. 
We then know that $3+b+c=9$ and $12+2b+c=19$ . So, subtracting the first from the second, we get that $9+b=10$. So $b=1$. Subbing $b=1$ into the first equation and solving for $c$, we find $c=5$. 
So we get to Azif00's numerator $3n^2+n+5$. 
Now for the sum. 
We can break it up. 
$$\sum_{n=1}^\infty \frac{3n^2+n+5}{n!}=\sum_{n=1}^\infty\frac{3n^2}{n!}+\sum_{n=1}^\infty\frac{n}{n!}+\sum_{n=1}^\infty\frac{5 }{n!}$$
Note for $e$ , $n$ must start with 0 and not 1, so we need to modify our terms. 
The third sum is $5(e-1/0!)=5(e-1).$
The second sum is of $1/(n-1)!$ with $n$ starting at 1. But the $(n-1)$ makes this start at 0. So the second sum is $e$. 
Now the first part of the sum is $\sum_{n=1}^\infty\frac{3n}{(n-1)!}$
We can change the index to k starting with k=0 and letting $n=k+1$. 
So the sum is $\sum_{k=0}^\infty \frac{3(k+1)}{k!}=\sum_{k=0}^\infty \frac{3k}{k!}+\sum_{k=0}^\infty\frac{3}{k!}$. The second sum is $3e$. 
The first term in the left hand sum is zero, so we can start it at k=1 instead and simply to sum over $3/(k-1)!$ But by similar arguments to the above, this is just $3e$. 
Put it all together we get $12e-5$
