# Soving Recurrence Relation

I have this relation

$u_{n+1}=\frac{1}{3}u_{n} + 4$

and I need to express the general term $u_{n}$ in terms of $n$ and $u_{0}$.

With partial sums I found this relation

$u_{n}=\frac{1}{3^n}u_{0} + 4\sum_{n=1}^n\frac{1}{3^n-1}$

But I also need to prove by mathematical induction that my $u_{n}$ is ok, but I have no idea how to do this. Can anyone please help me?

• It isn’t quite right: in $\sum_{n=1}^n\frac1{3^n-1}$ you’re using $n$ both as upper limit and as index. You want $\sum_{k=1}^n\frac1{3^k-1}$ or the like. – Brian M. Scott May 21 '13 at 18:16

$$u_{n+1}=\frac{1}{3}u_{n} + 4\implies 3u_n=u_n+12$$
$$\implies 3u_{n+2}=u_{n+1}+12$$
On subtraction, $$3u_{n+2}- 3u_{n+1}=u_{n+1}-u_n\implies 3u_{n+2}- 4u_{n+1}+u_n=0$$
$$u_{ n }=\frac { 1 }{ { 3 }^{ n } } u_{ 0 }+4\sum _{ k=0 }^{ n-1 }{ \frac { 1 }{ { 3 }^{ k } } } =\frac { 1 }{ { 3 }^{ n } } u_{ 0 }+4\left( \frac { { \left( 1/3 \right) }^{ n }-1\quad }{ 1/3-1 } \right) =\frac { 1 }{ { 3 }^{ n } } u_{ 0 }-6\left( \frac { 1 }{ { 3 }^{ n } } -1 \right)$$ $$u_{ n }=\frac { 1 }{ { 3 }^{ n } } \left(u_{ 0 }-6\right)+6$$