What is the sum of the first n terms of a geometric sequence and where does it come from?

Recall that a geometric progression is a sequence (a_n) defined as follows

 

a_1 = a

a_n = a * r^(n - 1) for all integers n > 1

where a and r are some fixed numbers and r /= 0 is called the common ratio of a sequence.

 

Let's denote S_n as the sum of the first n terms of this sequence.

 

Case r = 1

Then all the terms are equal to a so

S_n = na

 

Case r /= 1

Then we have

S_n = a_1 + a_2 + … + a_n = a + ar + … a*r^(n - 1) = a * (1 + r + … + r^(n – 1))

 

Now the formula for the difference between two n-th powers tells us that

r^n - 1 = r^n – 1^n = (r - 1) * (1 + r + … + r^(n-1))

and since r /= 1 we can divide both sides by r – 1 to have

1 + r + … + r ^(n - 1) = (r^n - 1) / (r – 1)

 

Finally, substituting this expression into the first equality we get

S_n = a * (r^n - 1) / (r – 1)

KN
Answered by Kamil N. Maths tutor

4795 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the indefinite integral of f(x)=(1-x^2)/(1+x^2)


How do you intergrate basic algebra?


A curve has equation y = 20x -x^(2) - 2x^(3). The curve has a stationary point at the point M where x = −2. Find the x coordinates of the other stationary point.


How do you find the coordinate of where two lines intersect?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning