<s>[INST] 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
 [/INST] 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
$$A_n=6a^n+\sum_{k=0}^{n-1}a^kb_{n-k}$$
$$=6(\frac 89)^n+\frac 65\sum_{k=0}^{n-1}(\frac 89)^k(\frac{20}{9})^{n-k}$$
$$=6(\frac 89)^n+\frac 65(\frac {20}{9})^n\sum_{k=0}^{n-1}(\frac 25)^k$$
$$=6(\frac{20}{9})^n\Bigl((\frac 25)^n+\frac 15\sum_{k=0}^{n-1}(\frac 25)^k\Bigr)$$
</s>