1. ## 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.

$

\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.

2. That's actually a telescoping sum. Do you know what a telescoping sum is?

3. 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.

4. 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.

$

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

$\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?

5. Thanks guys,

So essentially I'm left with

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