
Equation Proofs
I am having a lot of difficulty figuring this one out. The goal is to show that equations 1, 2, and 3, give the final simplified equation 4. I have always been more of a calculus person, and have trouble with these types of algebra problems.
http://rapidshare.com/files/449301420/GetEQ4.png
So far I have tried multiplying both S and Sigma together and attempted to put ga into the equation, which didn't really get me anywhere. Any help is appreciated!

I would plug Equations 2 and 3 into 4 and show that you get Equation 1, for an appropriate choice of N. Then, since all your steps are reversible, you've shown that which you were asked to show.

I plugged both 2 and 3 into the last equation and it simplified to this:
http://rapidshare.com/files/449308643/GetEQ4part2.png
The exponents still look incorrect and I missing something?

Nothing incorrect yet. Remember the laws of exponents, and simplify your expression a bit.

I was able to simplify the exponents a lot, and it looks better:
http://rapidshare.com/files/449313639/GetEQ4part3.png
I am still stuck with the ga term on the bottom though

You should have
$\displaystyle g_{a}^{n_{b}}$ on the bottom.
You now must choose the value of $\displaystyle N$ that makes it all work the same as Equation # 1.

Correct, I just saw that I forgot to raise the bottom ga to the nb power, that makes so much more sense!


Would it happen to be N=$\displaystyle n_b^2$, that way the ga term on the bottom would disappear completely leaving only the ga^nb on the top?

Not quite. You need to end up with $\displaystyle g_{a}^{n_{a}}$ on top. Right now, you have $\displaystyle g_{a}^{N}$ on top, and $\displaystyle g_{a}^{n_{b}}$ on the bottom. Use the laws of exponents. What must $\displaystyle N$ be?

N=$\displaystyle n_a+n_b$

There you go. I'd say you're done now, essentially. Just run your computations in reverse, and you can show that Equations 13 give you Equation 4.