Recurring Sequence with Exponent Working with recurring sequences and generating functions, I'm generally lost on solving a general expression of $a_n$ for any $n$ when the next part of the sequence, that is $a_{n+1}$, is in the form of an exponent, such that $a_n = a_{n-1} +k^{n-1}$, where k is some constant. I have  no clue on how to approach this problem.
I've solved the Fibonacci sequence by subtracting the two pervious terms and shifting the sequence, but it does not seem to work here.
I'm particularly working with $a_n = 2a_{n-1} + 5^{n-1}$, but the sequence expands extremely fast. The base case, $a_{0} = 1$.
Any help would be appreciated!

I love to telescope.
If
$a_n = ua_{n-1} + vc^{n}
$,
then
$\dfrac{a_n}{u^n}
= \dfrac{a_{n-1}}{u^{n-1}} + v(c/u)^{n}
$.
Let
$b_n = \dfrac{a_n}{u^n}$.
Then
$b_n
=b_{n-1}+vd^n
$
where
$d = c/u$.
Then
$b_n-b_{n-1}
=vd^n
$.
Summing,
$\begin{array}\\
b_m-b_0
&=\sum_{n=1}^m (b_n-b_{n-1})\\
&=\sum_{n=1}^m vd^n\\
&=v\dfrac{d-d^{m+1}}{1-d}\\
&=vd\dfrac{1-d^{m}}{1-d}\\
\end{array}
$
so
$\begin{array}\\
\dfrac{a_m}{u^m}
&=a_0+vd\dfrac{1-d^m}{1-d}\\
\text{or}\\
a_m
&=a_0u^m+\dfrac{vc}{u}u^m\dfrac{1-(c/u)^m}{1-c/u}\\
&=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\
&=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\
\end{array}
$
In  this case,
$u=2, c=5, v = \frac15, a_0 = 1
$
so
$a_m
= 2^m + \dfrac{2^m-5^m}{2-5}
= 2^m + \dfrac{5^m-2^m}{3}
$.
This can be rewritten as
$\begin{array}\\
a_m
&=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\
&=\dfrac{(u-c)a_0u^m+vc(u^m-c^m)}{u-c}\\
&=\dfrac{(a_0(u-c)+vc)u^m-vc^{m+1}}{u-c}\\
\end{array}
$
Again, we get
$=\dfrac{(a_0(u-c)+vc)u^m-vc^{m+1}}{u-c}
=\dfrac{(-3+1)2^m-5^{m}}{-3}
=\dfrac{2\cdot 2^m+5^{m}}{3}
$.
