Help with series needed please

**jayd1988** · November 9th 2010, 11:33 AM

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

Thanks!

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

**Chris L T521** · November 9th 2010, 11:42 AM

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?

**jayd1988** · November 9th 2010, 12:32 PM

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

**Chris L T521** · November 9th 2010, 12:36 PM

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

$\displaystyle\sum\limits_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$

**pickslides** · November 9th 2010, 12:41 PM

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

**jayd1988** · November 9th 2010, 01:58 PM

Thanks guys! Got it...

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