<s>[INST] How to solve this recurrence relation using generating functions: $a_n = 10 a_{n-1}-25 a_{n-2} + 5^n\binom{n+2}2$? How can we solve the following recurrence relation using GF?
$a_n = 10 a_{n-1}-25 a_{n-2} + 5^n {n+2 \choose 2}$ , for each $n>2, a_0 = 1, a_1 = 15$
I think that most of it is pretty straightforward. What really concerns me is this part
$5^n{n+2 \choose 2}$ 
Problem further analyzed
After trying to create generating functions in the equation we end up in this 
$\sum_{n=2}^\infty a_n x^n = 10 \sum_{n=2}^\infty {a_{n-1}} \sum_{n=2}^\infty a_{n-2} x^n + \sum_{n=2}^\infty {n+2 \choose 2}5^nx^n$
Lets take this part:
$ \sum_{n=2}^\infty {n+2 \choose 2}5^nx^n = \sum_{n=2}^\infty \frac{(n+1)(n+2)}{2} 5^nx^n $ 
then?
 [/INST] Hint for finding the GF. 
Note that $(n+1)(n+2)x^n=\frac{d^2}{dx^2}\left(x^{n+2}\right)$
and therefore 
$$\sum_{n=2}^\infty {n+2 \choose 2}5^nx^n = \frac{1}{2\cdot 5^2}\frac{d^2}{dx^2}\left(
\sum_{n=2}^\infty (5x)^{n+2}
\right).$$
Then recall that $\sum_{n=0}^{\infty}z^n=\frac{1}{1-z}$.
Moreover in your attempt it should be
$$\sum_{n=2}^\infty a_n x^n = 10 \sum_{n=2}^\infty {a_{n-1}}x^n-25 \sum_{n=2}^\infty a_{n-2} x^n + \sum_{n=2}^\infty {n+2 \choose 2}5^nx^n.$$
that is
$$\sum_{n=2}^\infty a_n x^n = 10x \sum_{n=1}^\infty {a_{n}}x^n-25x^2 \sum_{n=0}^\infty a_{n} x^n + \sum_{n=2}^\infty {n+2 \choose 2}5^nx^n.$$
Can you take it from here and find $f(x) =\sum_{n=0}^\infty a_n x^n$?
Alternative way for finding $a_n$ without the GF. 
The given recurrence is a non-homogeneous linear recurrence relation with constant coefficients with characteristic equation 
$$z^2-10z+25=(z-5)^2=0$$ 
and non-homogeneous term  $5^n {n+2 \choose 2}$ which is a second degree polynomial multiplied by a power of $5$, which is the root of multiplicity $2$ of the characteristic equation. 
Hence the general term of the recurrence with $a_0 = 1$, $a_1 = 15$ has the form 
$$a_n= 5^n(An^4+Bn^3+Cn+D)$$
where $A,B,C,D$ are real constant to be determined.
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