# Deriving a geometric series

1. Oct 23, 2006

### Elec68

I could use some help with this question:

Derive the geometric series representation of 1/(1-x) by finding a0, a1,
a2,... such that
(1-x)(a0+a1x+a2x^2+a3X^3+...)=1

Thank you.

2. Oct 23, 2006

### AKG

$$(1 - x)(a_0 + a_1x + a_2x^2 + a_3x^3 + \dots) = ? = \dots = ? = a_0 + (a_1-a_0)x + (a_2-a_1)x^2 + (a_3-a_2)x^3 + \dots$$

3. Oct 23, 2006

### Office_Shredder

Staff Emeritus
I recommend multiplying through by 1, then multiplying through by -x, then rearranging the terms

4. Oct 23, 2006

### mathman

How about (1-x)/(1-x)=1. Expand the denominator into a power series (binomial expansion of 1/(1-x)).

5. Oct 27, 2006

### fopc

If |x| <1, (1-x)*(1+x+x^2+x^3+x^4+...) = 1.
So your coefficients a_i = 1, for all i.

6. Oct 27, 2006

### Hurkyl

Staff Emeritus
(1) You (presumably) know how to multiply.
(2) You (presumably) know how to tell when two power series are equal.

There's no "trick" to this one -- you just do exactly what the equation suggests.