How to Find the Sum of Series with Alternating Sign and Variable Exponents

[Pr0grammer]

Find the sum of the series...

Homework Statement

Find the sum: $${1}^{2} - {2}^{2} + {3}^{2} - {4}^{2} + ... + [\left(-1\right)^{m-1}]{m^2}$$

and

Find the sum: $$x + {2x}^{2} + {3x}^{3} + ... + {nx}^{n}$$

Homework Equations

n/a

The Attempt at a Solution

For problem 1:
I tried splitting the series into 2 different series and adding them. $$\sum_{n=1}^{n} \left(2m-1\right)^{2} + \sum_{n=1}^{n} -\left(2m\right)^{2}$$
I need to figure out how to get the first sum to only apply to the odd terms and the second sum to only apply to the even terms.

For problem 2:
All I've managed to do is to split it into $${n}{x}{x}^{n-1}$$. I'm not really sure where to go after that, so any hints would be appreciated.

 

[micromass]

You could try to work out $$(2m-1)^2$$ and $$(2m)^2$$. This will make the sums easier...

 

[stevenb]

Homework Statement

Find the sum: $${1}^{2} - {2}^{2} + {3}^{2} - {4}^{2} + ... + [\left(-1\right)^{m-1}]{m^2}$$

and

Find the sum: $$x + {2x}^{2} + {3x}^{3} + ... + {nx}^{n}$$

Homework Equations

n/a

The Attempt at a Solution

For problem 1:
I tried splitting the series into 2 different series and adding them. $$\sum_{n=1}^{n} \left(2m-1\right)^{2} + \sum_{n=1}^{n} -\left(2m\right)^{2}$$
I need to figure out how to get the first sum to only apply to the odd terms and the second sum to only apply to the even terms.

For problem 2:
All I've managed to do is to split it into $${n}{x}{x}^{n-1}$$. I'm not really sure where to go after that, so any hints would be appreciated.

For the second problem, it looks like the following known trick will work.

Calculate $$S-aS$$ where $$S=\sum_{k=1}^{n}k a^k$$ and then rearrange to get an expression for $$S$$. You will find that the summation cancels out and leaves only the upper term, and there will also be another sum that is simpler and that you already should know. If somehow you don't know that simpler summation, you can apply the same trick to that.

If you've seen this trick before, you'll have no trouble. If you haven't, it's a little tricky the first time because you need to do an index substitution on the k-index. If you have trouble, try the following simpler problem first, using the same technique.

$$S=\sum_{k=0}^{n} a^k$$

 

[dickdick]

You might try factoring the first series as (1-2)(1+2)+(3-4)(3+4)+... if you want to be clever about it.