Don't understand this summation problem

$\displaystyle \sum\limits_{n = 1}^{2k} {{{\left( { - 1} \right)}^{n + 1}}n} = {\left( { - 1} \right)^{n + 1}}n$

and

$\displaystyle \sum\limits_{n = 1}^{2k} {{{\left( { - 1} \right)}^{n + 1}}} = {\left( { - 1} \right)^{n }}$

^Thanks to Plato for the Latex code

I just don't understand how to do these kind of things with a different letter on top. I know that I have to do something such as $\displaystyle (-1)^1 + (-1)^2 + ... + (-1)^{2k}$ for the second one but then how do you solve that?

Do I just ignore all the iterations until k? For example in the second one should I only look at $\displaystyle (-1)^{2k}*2k$? If so, shouldn't the answer be -2k since the first part will give you negative 1?

Sorry for the failed latex. I looked at the latex crash course but I still managed to screw everything up. Please can someone clean up my latex so I can at least understand how to use it properly?

Re: Don't understand this summation problem

Could you write the indexes of the sums? Thanks!

Re: Don't understand this summation problem

Sorry, the top of the first one is 2k, the bottom is n=1.

Next to it is (-1)^(n+1) *n

The second one has the same top and bottom but next to it (on the right) is (-1)^n

Re: Don't understand this summation problem

Quote:

Originally Posted by

**Mukilab** Sorry, the top of the first one is 2k, the bottom is n=1. Next to it is (-1)^(n+1) *n

The second one has the same top and bottom but next to it (on the right) is (-1)^n

Are these they?

$\displaystyle \sum\limits_{n = 1}^{2k} {{{\left( { - 1} \right)}^{n + 1}}n} = {\left( { - 1} \right)^{n + 1}}n$

$\displaystyle \sum\limits_{n = 1}^{2k} {{{\left( { - 1} \right)}^{n + 1}}} = {\left( { - 1} \right)^{n }}$?

You can click on "reply with quote" to see the LaTeX code.

Re: Don't understand this summation problem

Quote:

Originally Posted by

**Plato** Are these they?

$\displaystyle \sum\limits_{n = 1}^{2k} {{{\left( { - 1} \right)}^{n + 1}}n} = {\left( { - 1} \right)^{n + 1}}n$

$\displaystyle \sum\limits_{n = 1}^{2k} {{{\left( { - 1} \right)}^{n + 1}}} = {\left( { - 1} \right)^{n }}$?

You can click on "reply with quote" to see the LaTeX code.

Yes! Thank you very much, updated my OP.

Still waiting for an answer or help though :(