# Summation Help

• Aug 23rd 2009, 05:25 PM
Peleus
Summation Help
Hi all,

Just starting off learning summations, and I'm looking to find if there is an easier way to solve this problem.

$\displaystyle \sum (\frac{1}{k^2} - \frac{1}{(k-1)^2})$

The sum is k = 4, up to 24, sorry didn't know how to put it in for latex.

I know what it's asking, but surely there is a faster way than manually substituting in values and adding.

Any help is appreciated.
• Aug 23rd 2009, 05:36 PM
Krizalid
That's actually a telescoping sum. Do you know what a telescoping sum is?
• Aug 23rd 2009, 05:37 PM
Peleus
Nope, any chance you can point me in the right direction?

I'm guessing it has to do with the fact it's 1/16 - 1/9, then 1/25 - 1/16 etc etc, with the numbers shifting, but I'm not sure of the best way to deal with it.
• Aug 23rd 2009, 05:39 PM
Krizalid
• Aug 23rd 2009, 05:40 PM
skeeter
Quote:

Originally Posted by Peleus
Hi all,

Just starting off learning summations, and I'm looking to find if there is an easier way to solve this problem.

$\displaystyle \sum (\frac{1}{k^2} - \frac{1}{(k-1)^2})$

The sum is k = 4, up to 24, sorry didn't know how to put it in for latex.

I know what it's asking, but surely there is a faster way than manually substituting in values and adding.

Any help is appreciated.

$\displaystyle \sum_{k=4}^{24} \frac{1}{k^2} = \frac{1}{4^2} + \frac{1}{5^2} + ... + \frac{1}{24^2}$

$\displaystyle \sum_{k=4}^{24} -\frac{1}{(k-1)^2} = -\frac{1}{3^2} - \frac{1}{4^2} - ... - \frac{1}{23^2}$

sum these two series and what two terms remain?
• Aug 23rd 2009, 06:33 PM
Peleus
Thanks guys,

So essentially I'm left with

$\displaystyle \frac{1}{24^2} - \frac{1}{3^2}$