How to evaluate $ (0.3)^n\sum_{m=0}^n \left(\frac{0.8}{0.3}\right)^m $? Can someone please help me solve this sum:
$$
(0.3)^n\sum_{m=0}^n \left(\frac{0.8}{0.3}\right)^m u[n]
$$
where $u[n]$   means just that $n \ge 0$.
I keep getting $$-2(0.3^n -(0.8^(n+1))/(0.3))$$ but according to my textbook it is wrong.
I used the $$\sum_{i=m}^{n-1} a^i = \frac{a^m-a^n}{1-a}$$ forumla from which m becomes n+1. 
Thanks guys
I get this: $$ 2*(0.8)^(n+1) - (3/10)^n *3/5$$, and now I try to transform $$(3/10)^n$$ to 
$$(3/5)^n *(1/2)^n$$ and after that I can't do anything with $$(1/2)^n$$

$$
\sum_0^n a^n = \frac{1-a^{n+1}}{1-a}
$$
Substitute $a = \frac{0.8}{0.3} = \frac{8}{3} $ giving
$$\sum_0^n \left( \frac{8}{3} \right) ^m = \frac
{  1-\frac{8^{n+1}} {3^{n+1}}}
{1-\frac{8}{3}} = 
\frac{3}{5} \left( \frac{8^{n+1}} {3^{n+1}} -1 \right)
$$
Now multiply by $\left( \frac{3}{10} \right) ^ n$.  The powers of 3 cancel in the first term, and some powers of 2 cancel as well, and you are left with 
$$
\frac{ 2^{2n+3} }{5^{n+1}}  -
\frac{3^{n+1}} { 2^n \cdot 5^{n+1} }  $$ 
