<s>[INST] Sum of the multiplication of two geometric progressions I want to calculate the convolution of two geometric progressions by the convolution definition:
$$2^n*3^n=\sum_{k=0}^{n}f_1[k]f_2[n-k]=\sum_{k=0}^{n}2^k3^{n-k}=3^n\sum_{k=0}^n 2^k 3^{-k}=3^n\sum_{k=0}^n \bigg(\frac{2}{3}\bigg)^k=3^n\frac{1-\big(\frac{2}{3}\big)^n}{1-\frac{2}{3}}=3^n\frac{1-\big(\frac{2}{3}\big)^n}{\frac{1}{3}}=3^{n+1}\bigg(1-\Big(\frac{2}{3}\Big)^n\bigg)=3^{n+1}-2^n3^{-n}3^{n+1}=3^{n+1}-3\times 2^n$$
Since the sum is in k, the $3^n$ is out of the sum.
This result must be equivalent to that obtained by the Z transform:
$$2^n*3^n=F_1[z]F_2[z]=\frac{z}{z-2}\times \frac{z}{z-3}=G(z)$$
$$\frac{G(z)}{z}=\frac{z}{(z-2)(z-3)}=\frac{-2}{z-2}+\frac{3}{z-3}$$
Making the Z-transform I get:
$$g(n)=-2\times2^n+3\times 3^n=-2^{n+1}+3^{n+1}$$
I can not get the same result in both expressions. what did I do wrong?
 [/INST] Use 
\begin{eqnarray*}
\sum_{k=0}^{n} r^k = \frac{1-r^{n+1}}{1-r}
\end{eqnarray*}
with $r=\frac{2}{3}$. So we have 
\begin{eqnarray*}
2^n*3^n=\sum_{k=0}^{n}f_1[k]f_2[n-k]=\sum_{k=0}^{n}2^k3^{n-k}=3^n\sum_{k=0}^n 2^k 3^{-k} =3^n \frac{1-(\frac{2}{3})^{n+1}}{1-\frac{2}{3}} =3^{n+1}-2^{n+1}.
\end{eqnarray*}
</s>