# Series Question

1. Sep 29, 2005

### misogynisticfeminist

I've got a very tricky series question on hand,

I'm given $$\sum_{r=1}^{\infty} r^2 = \frac{1}{6} n(n+1)(2n+1)$$ and i'm asked to find $$\sum_{r=n+1}^{2n} 4r^2$$.

The change of limits and all is killing me ! any ideas on how to start? Thanks.

2. Sep 29, 2005

### HallsofIvy

There's an obvious typo in the "given" formula- there is no "n" on the left hand side. It should be
$$\sum_{r=1}^n r^2 = \frac{1}{6} n(n+1)(2n+1)$$

First, obvious point: ignore the "4" until you have done just $$\sum_{r=n+1}^{2n} r^2$$, then multiply by 4.

Can you use the "given" formula to find $$\sum_{r=1}^{2n} r^2$$?
Can you use the "given" formula to find $$\sum_{r=1}^{n} r^2$$?

What is the difference between them?

Last edited by a moderator: Sep 29, 2005
3. Sep 29, 2005

### misogynisticfeminist

OH yes ! thanks alot for the help, and esp. pointing out that typo. Am able to solve now.

: )