Sequences solutions with generating functions The task is to find $a_n$ (using generating functions preferably) when $a_0 = 1$ and for every $n\ge 0$ $$a_{n+1} = \frac{1}{3} \sum_{k=0}^{n} (8-5(-2)^k)\,a_{n-k}$$
I marked $b_n = \frac{1}{3}(8-5(-2)^n)$ and found the generating function $$g(x) = \sum_{n=0}^{\infty} \frac{1}{3}(8-5 (-2)^n)x^n = \frac{1+7x}{(1-x)(1+2x)}$$ 
I know that for suppose $t_n$ and $p_n$ are sequences with generating functions $f(x)$ and $p(x)$ then the generating function $h(x) = t(x)p(x)$ is for the sequence $$d_n = \sum_{k=0}^{n} t_{k} \, p_{n-k}$$
I know I need to use this here but I'm not really sure how.
 A: You are in the right track in invoking the convolution property of GF:
$$d_n = \sum_{k=0}^{n} t_{k} \, p_{n-k} \tag{1}$$
(Recall that this assumes we are using sequences indexed on the non-negative integers)
The problem here is that we have $a_k$ (our unknown) on both sides of the equation.
Then let define the sequence $$ u_n = a_{n+1}=\sum_{k=0}^{n} b_k \, a_{n-k} \tag{2}$$ for $n\ge0$
Let $A(x),B(x),U(x)$ be the GF of $a_k,b_k,u_k$.
Then, using $ u_n = a_{n+1} $ we get  (shifting property): $$U(x) = \sum_{n=0}^\infty x^n u_n= 
\sum_{n=0}^\infty x^n a_{n+1}=
\sum_{j=1}^\infty x^{j-1} a_j = x^{-1} (A(x)  -a_0) \tag{3}$$
Putting this together with the convolution property: $$U(x)=A(x)B(x)\tag{4}$$
and, given $a_0=1$, we get
$$A(x)=\frac{1}{1-x B(x)}$$
Can you go on from here?
A: Very good for the first part.
For continuing, the best approach is to include the starting conditions in the recurrence.
So you will write
$$
a_{\,n}  = \sum\limits_{0\, \le \,k\, \le \,n - 1} {b_{\,n - 1} \,a_{\,n - 1 - k} }  + \left[ {0 = n} \right]
$$
where $[P]$ denotes the Iverson bracket.
Then multiplying by $x^n$ and summing
$$
\eqalign{
  & f(x) = \sum\limits_{0\, \le \,n} {a_{\,n} \,x^{\,n} }  =   \cr 
  &  = \sum\limits_{0\, \le \,n} {\left( {\sum\limits_{0\, \le \,k\, \le \,n - 1} {b_{\,n - 1} \,a_{\,n - 1 - k} } } \right)\,x^{\,n} }  + \sum\limits_{0\, \le \,n} {\left[ {0 = n} \right]\,x^{\,n} }  =   \cr 
  &  = x\sum\limits_{0\, \le \,n} {\left( {\sum\limits_{0\, \le \,k\, \le \,n - 1} {b_{\,n - 1} \,a_{\,n - 1 - k} } } \right)\,x^{\,n - 1} }  + 1 =   \cr 
  &  = xg(x)f(x) + 1 \cr} 
$$
you get
$$
f(x) = {1 \over {1 - xg(x)}}
$$
