Simplifying the sequence defined by $A_0=6$ and $A_n=\frac{8}{9}A_{n-1}+\frac{6}{5}(\frac{20}{9})^n$, using sigma notation 
I have $A_0 = 6$ and
$$A_n = \left(\frac{8}{9}\right)A_{n-1} + \left(\frac{6}{5}\right) \left(\frac{20}{9}\right)^n$$
and I want to simplify it with sigma notation.

I have it up to
$$\begin{align}A_4 = &6 \left(\frac{8}{9}\right)^4 +  {\left(\frac{8}{9}\right)^3} \left(\frac{6}{5}\right) {\left(\frac{20}{9}\right)^{1}} + {\left(\frac{8}{9}\right)^2} \left(\frac{6}{5}\right){\left(\frac{20}{9}\right)^{2}} \\
&+ {\left(\frac{8}{9}\right)^1}\left(\frac{6}{5}\right){{\left(\frac{20}{9}\right)^{3}}} +  \left(\frac{6}{5}\right) {\left(\frac{20}{9}\right)^4} 
\end{align}$$ because I wanted to find the pattern
Apparently the answer is
$$4\left(\frac{8}{9}\right)^n + 2\left(\frac{20}{9}\right)^n$$ and the simplification into sigma notation is $$6\left(\frac{8}{9}\right)^n + \sum_{k=0}^{n}\left(\frac{8}{9}\right)^{n-k}\left(\frac{6}{5}\right)\left(\frac{20}{9}\right)^k$$
but I can seem to can seem to justify that. It's has to be some sort of algebraic manipulation that I did wrong but I don't see how the $k=0$ term works? To me, it seems that the sigma notation adds an extra term when $k=0$
(btw, this is based on the second-last and last pages of the PowerPoint presentation "Mathematics Geometry: Menger Sponge" (PPT link via ubc.ca) by the University of British Columbia Science and Mathematics Education Research Group.)
--
Personally, I think it's a typo since on Wolfram Alpha, if I use my $k=1$ then I get the right result but with $k=0$ it doesn't work
 A: (I have done this type of problem
so often,
I could almost do it in my sleep.
However,
each time I do it a little differently
and from scratch,
so here is the
latest version.)
If
$a_n
=ua_{n-1}+b_n
$
then,
dividing by $u^n$,
$\dfrac{a_n}{u^n}
=\dfrac{ua_{n-1}}{u^n}+\dfrac{b_n}{u^n}
=\dfrac{a_{n-1}}{u^{n-1}}+\dfrac{b_n}{u^n}
$.
Letting
$c_n = \dfrac{a_n}{u^n}$,
this becomes
$c_n
=c_{n-1}+\dfrac{b_n}{u^n}
$
so
$c_n-c_{n-1}
=\dfrac{b_n}{u^n}
$.
The left side telescopes
when summed so that
$c_m-c_0
=\sum_{n=1}^m (c_n-c_{n-1})
=\sum_{n=1}^m \dfrac{b_n}{u^n}
$
or
$\dfrac{a_m}{u^m}
=a_0+\sum_{n=1}^m \dfrac{b_n}{u^n}
$
or
$a_m
=u^ma_0+\sum_{n=1}^m u^{m-n}b_n
$.
If
$b_n=r s^n$
(as in this problem),
$\begin{array}\\
a_m
&=u^ma_0+\sum_{n=1}^m u^{m-n}rs^n\\
&=u^ma_0+ru^m\sum_{n=1}^m (s/u)^n\\
&=u^ma_0+ru^m\sum_{n=1}^m t^n
\qquad t = s/u\\
&=u^ma_0+ru^m\dfrac{t-t^{m+1}}{1-t}
\qquad\text{if } t \ne 1\\
&=u^ma_0+ru^m\dfrac{s/u-(s/u)^{m+1}}{1-s/u}\\
&=u^ma_0+r\dfrac{su^{m-1}-s^{m+1}/u}{1-s/u}\\
&=u^ma_0+r\dfrac{su^{m}-s^{m+1}}{u-s}\\
&=u^ma_0+rs\dfrac{u^{m}-s^{m}}{u-s}\\
&=u^m(a_0+\dfrac{rs}{u-s})-rs\dfrac{s^{m}}{u-s}\\
&=u^m(a_0+v)-vs^m
\qquad v=\dfrac{rs}{u-s}\\
\\
&=u^ma_0+ru^mm
\qquad\text{if } t = 1\\
&=u^m(a_0+rm)\\
\end{array}
$
If
$a_0=6,
u=\dfrac89, 
r=\dfrac65,
s=\dfrac{20}{9}
$
then
$u-s
=-\dfrac{12}{9}
=-\dfrac43
$,
$rs
=\dfrac83
$,
$v
=-2
$
so that
$\begin{array}\\
a_n
&=u^m(a_0+v)-vs^m\\
&=(8/9)^m(6-2)+2(20/9)^m\\
&=\dfrac{4\cdot 8^n+2\cdot 20^n}{9^n}\\
\end{array}
$
A: For $ n\ge 1$,
$$A_n=aA_{n-1}+b_n$$
$$aA_{n-1}=a^2A_{n-2}+ab_{n-1}$$
$$a^2A_{n-2}=a^3A_{n-3}+a^2b_{n-2}$$
$$a^kA_{n-k}=a^{k+1}A_{n-k-1}+a^kb_{n-k}$$
...
$$a^{n-1}A_1=a^nA_0+a^{n-1}b_1$$
the sum gives
$$A_n=6a^n+\sum_{k=0}^{n-1}a^kb_{n-k}$$
$$=6(\frac 89)^n+\frac 65\sum_{k=0}^{n-1}(\frac 89)^k(\frac{20}{9})^{n-k}$$
$$=6(\frac 89)^n+\frac 65(\frac {20}{9})^n\sum_{k=0}^{n-1}(\frac 25)^k$$
$$=6(\frac{20}{9})^n\Bigl((\frac 25)^n+\frac 15\sum_{k=0}^{n-1}(\frac 25)^k\Bigr)$$
