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?
Thanks in advance

$$u_{ n }=\frac { 1 }{ { 3 }^{ n } } u_{ 0 }+4\sum _{ k=0 }^{ n-1 }{ \frac { 1 }{ { 3 }^{ k } }  } =\frac { 1 }{ { 3 }^{ n } } u_{ 0 }+4\left( \frac { { \left( 1/3 \right)  }^{ n }-1\quad  }{ 1/3-1 }  \right) =\frac { 1 }{ { 3 }^{ n } } u_{ 0 }-6\left( \frac { 1 }{ { 3 }^{ n } } -1 \right) $$ $$u_{ n }=\frac { 1 }{ { 3 }^{ n } } \left(u_{ 0 }-6\right)+6$$
See geometric series
