How to find the closed form of $f(n) = 9^k \times (-56) + f(n-1)$ I have to find the closed form for 
$$T(n) = \begin{cases} 
2 , &\text{ if } n=0 \\
9T(n-1) - 56n + 63, &\text{ if } n > 1
\end{cases}$$
I used the repeated substitution method and I found that the pattern for the coefficient of n is equal to the following:
$$f(1) = -56$$
$$f(n) = 9^{n-1} \times (-56) + f(n-1)$$
I tried to find the closed form of $f(n) = 9^{n-1} \times (-56) + f(n-1)$, but it just got more and more confusing. I believe it may be a series of some sort. Is there a way to find a closed form for this?
Thank you!

Usually, these problems are solved using induction. But induction requires you to already 'know' (or guess) the answer. Using generating functions we can solve the problem without having to ever guess:
$$t(n) = 9t(n-1) - 56n + 63$$
$$t(0) = 2$$
Now suppose we write:
$$T(x) = \sum_{n=0}^\infty t(n)x^n$$
We know:
$$T(x) = 2 + \sum_{n=1}^\infty (9t(n-1) - 56n + 63)x^n$$
$$T(x) = 2 +9\sum_{n=1}^\infty t(n-1)x^n -56\sum_{n=1}^\infty nx^n+ 63 \sum_{n=1}^\infty x^n$$
$$T(x) = 2 +9\sum_{n=0}^\infty t(n)x^{n+1} -56\frac{x}{(1-x)^2}+ 63 \frac{x}{1-x}$$
$$T(x) = 2 +9xT(x) -56\frac{x}{(1-x)^2}+ 63 \frac{x}{1-x}$$
$$T(x) = 2\frac{1}{1-9x} -56\frac{x}{(1-9x)(1-x)^2}+ 63 \frac{x}{(1-9x)(1-x)}$$
$$T(x) = 2\frac{1}{1-9x} +\frac{-56x + 63x(1-x)}{(1-9x)(1-x)^2}$$
$$T(x) = 2\frac{1}{1-9x} +\frac{7x(1-9x)}{(1-9x)(1-x)^2}$$
$$T(x) = 2\frac{1}{1-9x} +7\frac{x}{(1-x)^2}$$
$$T(x) = 2\sum_{n=0}^\infty 9^nx^n + 7\sum_{n=0}^\infty nx^n$$
$$T(x) = \sum_{n=0}^\infty \Big (2\cdot 9^n  + 7n\Big)x^n = \sum_{n=0}^\infty t(n)x^n$$
Therefore, $t(n) = 2\cdot 9^n + 7n$.
