The sum of the first *n* terms of the geometric sequence, in expanded form, is as follows:

*a* + *ar* + *ar^*2 + *ar^*3 + ... + *ar^n*–2 + *ar^n*–1

Polynomials written with their terms in descending order. Reversing the order of the summation above to put its terms in descending order, we get a series expansion of:

*ar^n*–1 + *ar^n*–2 + ... + *ar^*3 + *ar^*2 + *ar* + *a*

We can take the common factor of "*a*" out front:

*a*(*r^n*–1 + *r^n*–2 + ... + *r^*3 + *r^*2 + *r* + 1)

A basic property of polynomials is that if you divide *x^(n* – 1) by *x* – 1, you'll get:

*x^n*–1 + *x^n*–2 + ... + *x^*3 + *x^*2 + *x* + 1

This gives:

x^{n-1} + ... + x^2 + x + 1 = {x^n - 1}/{x - 1}

If we reverse both subtractions in the fraction above, we will obtain the following equivalent equation:

x^{n-1} + ... + x^2 + x + 1 = {1 - x^n}/{1 - x}

Applying the above to the geometric summation (by using *r *instead of *x*), we get:

ar^{n-1} + ar^{n-2} + ... + ar^2 + ar + a

a(r^{n-1} + r^{n-2} + ... + r^2 + r + 1)

= a{1 - r^n}/{1 - r} , proven