Finding closed forms for ∑kx^k and ∑k²x^k by differentiating a geometric series

[Chewy0087]

Homework Statement

Given that

$$\sum^{n}_{k=0} x^{k}= \frac{1-x^{n+1}}{1-x}$$

Obtain a similar result for ;

$$\sum^{n}_{k=1} kx^{k}$$
$$\sum^{n}_{k=1} k^{2}x^{k}$$

The Attempt at a Solution

Hey, well basically my trouble with this question stems from the manipulation of the limits and the effects that it has on the series itself;

For the first one, I differentiated giving me;

$$\sum^{n}_{k=1} kx^{k-1}= \frac{x^{n+1}-x^{n}(n+1)+1}{(1-x)^{2}}$$

Obviously when k<0 it doesn't hold as it would be 0, so I changed it to 1 from 0, and I can then reduce the lower index to 0 to make it into a better form.

However I'm just confused as to how to manipulate the index, and as to whether it would have any effect on the initial RHS equation. I don't seem to be getting the right answer in the way I've done above.

I have looked this up but no-where seems to be very helpful about it...

Thanks

 

[fzero]

Manipulating the index in a sum is as easy as changing variables. If k starts at k=1, then s=k-1 starts at 0. So

$$\sum^{n}_{k=1} kx^{k-1} = \sum_{s=0}^{n-1} (s+1) x^s.$$

 

[Chewy0087]

I have a feeling that it's something to do with discarding the k=0 and so changing to to k=1, and then changing the index back to 0 and doing the same? But that feels like such a cop-out.

 

[Chewy0087]

Manipulating the index in a sum is as easy as changing variables. If k starts at k=1, then s=k-1 starts at 0. So

$$\sum^{n}_{k=1} kx^{k-1} = \sum_{s=0}^{n-1} (s+1) x^s.$$

I feel so stupid, I wasn't reducing the upper limit >.<.

Thanks bud