# Thread: Summing a geometric series

1. ## Summing a geometric series

Hi all. I have been asked to sum this series:

1+4x+7x^2+10x^3+...+(3N-2)x^N

and also asked what the sum is to infinity when |x|< 1

I have all of the formulae (dodgy spelling ) for summing a series and sum to infinity but this one has me stumped; mainly because at the back of the text book the given anser is:

[3/(1-x)^2]-[2/(1-x)]

I just don't know how to get this answer. It's caused a bit of a mental block for me and his becoming quite annoying! Any help on working through to get this answer would be really helpful and much appreciated. Thanks in advance!

2. Originally Posted by morrey
Hi all. I have been asked to sum this series:

1+4x+7x^2+10x^3+...+(3N-2)x^{N-1}

and also asked what the sum is to infinity when |x|< 1

I have all of the formulae (dodgy spelling ) for summing a series and sum to infinity but this one has me stumped; mainly because at the back of the text book the given anser is:

[3/(1-x)^2]-[2/(1-x)]

I just don't know how to get this answer. It's caused a bit of a mental block for me and his becoming quite annoying! Any help on working through to get this answer would be really helpful and much appreciated. Thanks in advance!
You have a typo (corrected in red).

You have $\displaystyle \sum_{i=1}^{N} (3i - 2) x^{i-1} = 3 \sum_{i=1}^{N} i x^{i-1} - 2 \sum_{i=1}^{N} x^{i-1}$.

You know $\displaystyle \sum_{i=1}^{N} x^i$ and $\displaystyle \sum_{i=1}^{N} x^{i-1}$.

Differentiate $\displaystyle \sum_{i=1}^{N} x^i$ to get what $\displaystyle \sum_{i=1}^{N} i x^{i-1}$ is.

3. fantastic, thanks very much. Just what I needed, and as always looking back I wonder how i got stuck! thanks again

4. Hello, morrey!

I don't agree with their answer for the sum of the infinite series.

Find the sum: .$\displaystyle 1+4x+7x^2+10x^3+\hdots$

$\displaystyle \begin{array}{ccccc}\text{We have:} & S &=& 1 + 4x + 7x^2 + 10x^3 + 13x^4 + \hdots & {\color{blue}[1]}\\ \text{Multiply by }x: & xS &=& \qquad\; x \;+ 4x^2 + \;7x^3 + 10x^4 + \hdots & {\color{blue}[2]}\end{array}$

Subtract [1] - [2]: .$\displaystyle S - xS \;=\;1 + 3x + 3x^2 + 3x^3 + \hdots$

$\displaystyle \text{We have: }(1-x)S \;=\;1 + 3x\underbrace{\left(1 + x + x^2 + x^3 + \hdots\right)}_{\text{geometric series}}$

. . The geometric series has the sum: .$\displaystyle \frac{1}{1-x}$

Hence: .$\displaystyle (1-x)S \:=\:1 + 3x\left(\frac{1}{1-x}\right) \;=\; \frac{1+2x}{1-x}$

. . Therefore: .$\displaystyle \boxed{S \;=\;\frac{1+2x}{(1-x)^2}}$

5. Originally Posted by Soroban

$\displaystyle \text{We have: }(1-x)S \;=\;1 + 3x\underbrace{\left(1 + x + x^2 + x^3 + \hdots\right)}_{\text{geometric series}}$

. .
Where you have factorised and brought out (1+3x), can you do that? Because if you multiplied out those brackets you wouldn't get what you factorised in the first place... I can see your working through though. Just that one point that has confused me a little.

EDIT: also @mr.fantastic i'm not sure it was a typo, that is exactly as written in the text book, unless of course it's typed incorrectly in there! Also, where you have seperated the two sums. Is it possible to take the right summation and change it into the form where you start at i=0 up to i and then x^i-1 becomes just x^i. Because then from there I know what that sum is but can you then subtract the two different answers although they are 'summed' in a slightly different form? THanks for your help so far

EDIT 2: @soropan sorry about that just realised how you have written it. Makes sense now and I agree with what you have given me. Thankyou very much for your quick and friendly response! One point to make actually, could you not just work out the sum for 1+3x+3x^2.... instead of factorising?

6. Originally Posted by morrey
[snip]
EDIT: also @mr.fantastic i'm not sure it was a typo, that is exactly as written in the text book, unless of course it's typed incorrectly in there! Also, where you have seperated the two sums. Is it possible to take the right summation and change it into the form where you start at i=0 up to i and then x^i-1 becomes just x^i. Because then from there I know what that sum is but can you then subtract the two different answers although they are 'summed' in a slightly different form? THanks for your help so far

[snip]
If you substitute values of N into $\displaystyle (3N-2)x^N$ you will see why I think there's a typo ....

$\displaystyle \sum_{i=1}^{N} x^{i-1} = \sum_{i=0}^{N-1} x^{i}$.

7. Yes, you are correct. It seems that all this amount of confusion on my part partially comes down to a damn silly text book with an incorrect answer and question! Oh well, thanks to you both for your help and explanations!