Generating Function for a Recurrence: $a_k = 3a_{k−1} + 4$

Question

Use generating functions to solve the recurrence relation $a_k = 3a_{k−1} + 4$ with the initial condition $a_0 = 1$.

I have done my work until $(4-x)\sum_\limits0^\infty (n+1)x^n$

and got stuck here.

Answer 1

From $$ a_k=3a_{k-1}+4, \quad k=1,2,3,\ldots $$ setting $ f(x):=\sum_{k=0}^\infty a_kx^k$, you get $$ \sum_{k=1}^\infty a_kx^k=3x\sum_{k=1}^\infty a_{k-1}x^{k-1}+4\sum_{k=1}^\infty x^k $$ $$ f(x)-a_0=3xf(x)+4\frac{x}{1-x} $$ giving, with $a_0=1$,

$$ f(x)=\frac{3}{1-3x}-\frac{2}{1-x}.\tag1 $$

Then expand $(1)$ for $|x|<1/3$.

Can you take it from here?