Help with evaluating a sum

Hello everyone,

I'm not sure if this is the best category for my post, but i will try asking anyways. I've spent quite a bit of time trying to figure it out.

Basically, the problem involves calculating the output of a system to a given input and impulse response using convolution. That's probably the hardest part of the question, which I am able to do.

I get the output y[n] = summation(from k=0 to k=n) ((1/2)^k)*(1/3)^(n-k)

So i know that (1/3)^n can be brought outside the summation, and that

((1/2)^k)*(1/3)^(-k) = (3/2)^k, so the question can be simplified to

y[n]=((1/3)^n)*summation(k=0 to k=n) (3/2)^k

I know the formula for evaluating a sum from 0 to n, but for some reason i can't simplify it to the answer in the textbook. The answer i need is the following:

y[n]=2[((3/2)^(n+1))-1]

Any help is greatly appreciated.

The question can be seen here: http://www.site.uottawa.ca/~jpyao/co..._Fall_2004.pdf

Question #4

Re: Help with evaluating a sum

$\displaystyle \left(\frac{1}{3}\right)^n \sum_{k=0}^n \left(\frac{3}{2}\right)^k$

sum of the finite geometric series ...

$\displaystyle S_n = \frac{a_0 - r^{n+1}}{1 - r}$

$\displaystyle S_n = \frac{1 - \left(\frac{3}{2}\right)^{n+1}}{1 - \left(\frac{3}{2}\right)}$

$\displaystyle S_n = \frac{1 - \left(\frac{3}{2}\right)^{n+1}}{-\frac{1}{2}}$

$\displaystyle S_n = 2\left[\left(\frac{3}{2}\right)^{n+1} - 1\right]$

so ...

$\displaystyle \left(\frac{1}{3}\right)^n \sum_{k=0}^n \left(\frac{3}{2}\right)^k = \left(\frac{1}{3}\right)^n \cdot 2\left[\left(\frac{3}{2}\right)^{n+1} - 1\right]$

did the book leave off the $\displaystyle \left(\frac{1}{3}\right)^n$ or did you?

Re: Help with evaluating a sum

Wow, amazing how simple that was. I was way overcomplicating it.

The book left off the (1/3)^n apparently. Thanks for clarifying it for me. Still can't believe i spent over an hour trying to figure it out!