Hello, can somebody please explain the intermediate steps I take to solve the following?
Thanks!
$\displaystyle \displaystyle \frac{2}{M} \sum_{n=1}^{\frac{M}{2}} (2n-1)^2 A^2 = \frac{M^2-1}{3} A^2$
First off, the $\displaystyle A^2$ can be pulled out. So we have $\displaystyle \displaystyle\frac{2A^2}{M}\sum\limits_{n=1}^{\fra c{M}{2}}(2n-1)^2$. Take $\displaystyle k=2n-1$. Then we can rewrite the sum as $\displaystyle \displaystyle\frac{2A^2}{M}\sum\limits_{k=1}^{M-1}k^2$. Now apply the formula for $\displaystyle \displaystyle\sum_{k=1}^n k^2$ and simplify. From there, you should get the desired result.
Does this make sense?