# Thread: Sum of Infinite Series

1. ## Sum of Infinite Series

Just came across this problem in one of my past papers (which annoyingly dont come with answers) :

Find as a function of x the sum of the series

x + 3x^2 + 9x^3 + 27x^4 + ((-1)^n)*3^n*x^(n+1) + ....

I have solved similar problems before with binomial and geometric series expressions but I'm not sure how I'd go about this particular series.

Any help?

2. Originally Posted by thepars
Just came across this problem in one of my past papers (which annoyingly dont come with answers) :

Find as a function of x the sum of the series

x + 3x^2 + 9x^3 + 27x^4 + ((-1)^n)*3^n*x^(n+1) + ....

I have solved similar problems before with binomial and geometric series expressions but I'm not sure how I'd go about this particular series.

Any help?
I think this is what you mean you have a $(-1)^n$ in your general term but you sign does not alternate

$\sum_{n=0}^{\infty}(-1)^n3^{n}x^{n+1} =x\cdot \sum_{n=0}^{\infty}(-1)^n3^{n}x^{n}=x\cdot \sum_{n=0}^{\infty}(-3x)^n$

This is a geometric series so we get

$x\cdot \frac{1}{1+3x}=\frac{x}{1+3x}$

3. Originally Posted by TheEmptySet
I think this is what you mean you have a $(-1)^n$ in your general term but you sign does not alternate

$\sum_{n=0}^{\infty}(-1)^n3^{n}x^{n+1} =x\cdot \sum_{n=0}^{\infty}(-1)^n3^{n}x^{n}=x\cdot \sum_{n=0}^{\infty}(-3x)^n$

This is a geometric series so we get

$x\cdot \frac{1}{1+3x}=\frac{x}{1+3x}$
Wow it was so obvious I think I'm just having a bad day maths-wise.

Thanks