Recurrence Relations Closed Form So, the question is to derive the closed form solution to the recurrence relation $$T(n) = 3T(n-1) + 5,\hspace{5mm} T(0) = 0.$$
$\begin{align}T(n) &= 3T(n-1)+5
    \\&= 3(3T(n-2)+5)+5
     \\&= 3(3(3T(n-3)+5)+5)+5\end{align}$
I'm just struggling to go from here. I understand that it's $3$ to the power of something but I'm lost on what I have to do after this. 

Calculating the first few values: $0,5,20,65,200$ , dividing by $5$ this is $0,1,4,13,40$ , multiplying by $2$ we get $0,2,8,26,80$ adding $1$ we get $1,3,9,27,81$. Otherwise you can use generating functions:
Let $A=\sum\limits_{n=0}^\infty t(n)x^n$.
Then $A=3xA+5\sum\limits_{n=1}^\infty x^n$ (since $f(0)=0$).
Then $(1-3x)A=\frac{5}{1-x}-5$ so $A=\frac{5}{(1-3x)(1-x)}-\frac{5}{1-3x}=\frac{5}{2(x-1)}-\frac{15}{2(3x-1)}+\frac{5}{3x-1}=\frac{5}{2(x-1)}-\frac{5}{2(3x-1)}=\frac{1}{2}(\sum\limits_{n=0}^\infty 5\cdot 3^{n} -5)$
therefore $f(n)=\frac{5(3^{n}-1)}{2}$
