determine x the sum of the Geometric series

[Louise Johnson]

Question: The sum of the infinite geometric series

\(\displaystyle \L\\(x - 1) + x - 1)^2 + (x - 1)^3 + .........is\:\frac{1}{3}\) Determine the value of x

What I have so far:
a=(x-1)
r=(x-1)

using the formula

\(\displaystyle S = \frac{a}{{1 - r}}\)


\(\displaystyle \frac{1}{3} = \frac{{x - 1}}{{1 - (x - 1)}}\)

That is as far as I have got on this geometric series
Any help out there?
Thank you
Louise

 

[galactus]

That's right. Solve for x.

\(\displaystyle \L\\\frac{x-1}{2-x}=\frac{1}{3}\)

x=5/4, 5/4-1=1/4

The series is \(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{1}{4^{n}}=\frac{1}{3}\)

 

[Louise Johnson]

Thank you Galactus. I was unsure I was going in the right direction. Gonna spend some time reviewing the solution.
Thank you
Louise