1. ## Help with series needed please

Hello, can somebody please explain the intermediate steps I take to solve the following?

Thanks!

$\displaystyle \frac{2}{M} \sum_{n=1}^{\frac{M}{2}} (2n-1)^2 A^2 = \frac{M^2-1}{3} A^2$

2. Originally Posted by jayd1988
Hello, can somebody please explain the intermediate steps I take to solve the following?

Thanks!

$\displaystyle \frac{2}{M} \sum_{n=1}^{\frac{M}{2}} (2n-1)^2 A^2 = \frac{M^2-1}{3} A^2$
First off, the $A^2$ can be pulled out. So we have $\displaystyle\frac{2A^2}{M}\sum\limits_{n=1}^{\fra c{M}{2}}(2n-1)^2$. Take $k=2n-1$. Then we can rewrite the sum as $\displaystyle\frac{2A^2}{M}\sum\limits_{k=1}^{M-1}k^2$. Now apply the formula for $\displaystyle\sum_{k=1}^n k^2$ and simplify. From there, you should get the desired result.

Does this make sense?

3. Originally Posted by Chris L T521
First off, the $A^2$ can be pulled out. So we have $\displaystyle\frac{2A^2}{M}\sum\limits_{n=1}^{\fra c{M}{2}}(2n-1)^2$. Take $k=2n-1$. Then we can rewrite the sum as $\displaystyle\frac{2A^2}{M}\sum\limits_{k=1}^{M-1}k^2$. Now apply the formula for $\displaystyle\sum_{k=1}^n k^2$ and simplify. From there, you should get the desired result.

Does this make sense?
Many thanks for your help. Sorry, which formula do I use for $\displaystyle\sum_{k=1}^n k^2$ — almost there

4. Originally Posted by jayd1988
Many thanks for your help. Sorry, which formula do I use for $\displaystyle\sum_{k=1}^n k^2$ — almost there
$\displaystyle\sum\limits_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$

5. $\displaystyle\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

6. Thanks guys! Got it...