How Do You Derive the Geometric Series for 1/(1-x)?

[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.

 

[AKG]

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

 

[Office_Shredder]

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

 

[mathman]

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

 

[fopc]

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

 

[Hurkyl]

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

(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.