How to evaluate $ (0.3)^n\sum_{m=0}^n \left(\frac{0.8}{0.3}\right)^m $?

Question

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 $$(1/2)^n$$

Answer 1

You get $\sum_{k=1}^n x^k = \frac{x^{n+1}-1}{x-1}$, so $$\sum_{m=0}^n (\frac{8}{3})^m = \frac{1-(8/3)^{n+1}}{1-8/3}=-0.6( 1-(8/3)^{n+1}) =-2\cdot 0.3( 1-(8/3)^{n+1}), $$ hence $$0.3^n\sum_{m=0}^n (\frac{8}{3})^m = -2 \cdot 0.3\cdot 0.3^{n}(1- (0.8/0.3)^{n+1} )$$ $$= -2 \cdot 0.3^{n+1}(1- (0.8/0.3)^{n+1} =-2(0.3^{n+1}-0.8^{n+1}).$$

Answer 2

$$ \sum_0^n a^n = \frac{1-a^{n+1}}{1-a} $$ 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 $$ \frac{ 2^{2n+3} }{5^{n+1}} - \frac{3^{n+1}} { 2^n \cdot 5^{n+1} } $$