How Do You Calculate the Sum of Squares from \(n+1\) to \(2n\)?

[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.

 

[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?

 

[misogynisticfeminist]

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

: )