How to solve this recurrence relation What are some ways to solve this recurrence relation:
$a(n+1)=2 a(n) - a(n-1) -1, \text{ with }a(0)=0, a(10)=0?$
I tried to first convert this inhomogeneous equation into a homogeneous one following Wikipedia:

$b_{n}=Ab_{n-1}+Bb_{n-2}+K \,$ can be
  converted into homogeneous form as
  follows: The steady state is found by
  setting $b_n = b_{n−1} = b_{n−2} = b^{*}$ to
  obtain $b^{*} = \frac{K}{1-A-B}. \, $.
  Then the non-homogeneous recurrence
  can be rewritten in homogeneous form
  as $[b_n -b^{*}]=A[b_{n-1}-b^{*}]+B[b_{n-2}-b^{*}], \, $

But in the recurrence relation given at the beginning of this post, the denominator in $b^{*} = \frac{K}{1-A-B} \, $ is $1-2+1=0$. So how can I solve the recurrence relation? Thanks in advance! 
PS: Are there some books or websites with some summary of various ways to solve recurrence relation?

Generating functions to the rescue. Define $A(z) = \sum_{n \ge 0} a(n) z^n$,
multiply the shifted by 2 recurrence by $z^n$ and sum over $n \ge 0$. Recognizing:
\begin{align}
\sum_{n \ge 0} a(n + 1) z^n &= \frac{A(z) - a(0)}{z} \\
\sum_{n \ge 0} a(n + 2) z^n &= \frac{A(z) - a(0) - a(1) z}{z^2} \\
\sum_{n \ge 0} z^n          &= \frac{1}{1 -z}
\end{align}
gives, using the unknown $a(1)$:
$$
\frac{A(z) - a(1) z}{z^2} = 2 \frac{A(z)}{z} - A(z) - \frac{1}{1 - z}
$$
From this maxima gets the partial fraction expansion:
$$
A(z) = - \frac{1 + a(1)}{1 - z} + \frac{2 + a(1)}{(1 - z)^2} - \frac{1}{(1 - z)^3}
$$
From here, using the generalized binomial theorem:
$$
a(n) = - (1 + a(1)) + (n + 1)(2 + a(1)) - \frac{(n + 1) (n + 2)}{2}
     = - \frac{n^2 - (2 a(1) + 1) n}{2}
$$
As $a(10) = 0$, this gives $a(1) = 9 / 2$, and thus:
$$
a(n) = \frac{n (10 - n)}{2}
$$
