Geometric Series Formula and Sum

A geometric series is what you get when you add up the terms of a geometric sequence. In a geometric sequence, each term is found by multiplying the one before it by a fixed number called the common ratio, usually written as \(r\). So if the first term is \(a_1\), the sequence looks like

\(a_1,\ a_1r,\ a_1r^2,\ a_1r^3,\ \dots\)

and the geometric series is simply the sum of those terms:

\(a_1 + a_1r + a_1r^2 + a_1r^3 + \dots + a_1r^{n-1}\)

This page focuses on the sum of a finite number of terms. If you want to add up infinitely many terms of a geometric series, that's a related idea covered in infinite geometric series.

Instead of adding every term one by one, there's a shortcut formula for the sum of the first \(n\) terms of a geometric series:

Here, \(S_n\) is the sum of the first \(n\) terms, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms being added. If \(r = 1\), every term is the same, so the sum is simply \(S_n = n \times a_1\).

The formula isn't just a rule to memorize, it comes from a neat trick. Start with the sum

\(S_n = a_1 + a_1r + a_1r^2 + \dots + a_1r^{n-1}\)

Now multiply every term by \(r\):

\(rS_n = a_1r + a_1r^2 + \dots + a_1r^{n-1} + a_1r^n\)

Subtract the second equation from the first. Almost every term cancels out, leaving only the first term of \(S_n\) and the last term of \(rS_n\):

\(S_n - rS_n = a_1 - a_1r^n\)

Factor both sides:

\(S_n(1 - r) = a_1(1 - r^n)\)

Divide both sides by \(1 - r\) (as long as \(r \ne 1\)) to get the formula:

\(S_n = \dfrac{a_1(1 - r^n)}{1 - r}\)

Find the sum of the first 5 terms of the geometric series with first term \(a_1 = 3\) and common ratio \(r = 2\).

The terms are \(3, 6, 12, 24, 48\). Plug the values into the formula:

\(S_5 = \dfrac{3(1 - 2^5)}{1 - 2} = \dfrac{3(1 - 32)}{-1} = \dfrac{3(-31)}{-1} = 93\)

You can check this by watching how the sum builds up term by term:

Find the sum of the first 4 terms of the geometric series with \(a_1 = 100\) and \(r = \dfrac{1}{2}\).

The terms are \(100, 50, 25, 12.5\). Using the formula:

\(S_4 = \dfrac{100\left(1 - \left(\frac{1}{2}\right)^4\right)}{1 - \frac{1}{2}} = \dfrac{100(1 - 0.0625)}{0.5} = \dfrac{100(0.9375)}{0.5} = 187.5\)

Notice that when \(0 < r < 1\), the terms shrink instead of grow, but the formula works exactly the same way.

Because a geometric series is just a sum of terms following a pattern, it can also be written compactly using sigma notation. The first example above can be written as

\(\displaystyle\sum_{k=1}^{5} 3(2)^{k-1} = 93\)

This says: plug \(k = 1, 2, 3, 4, 5\) into \(3(2)^{k-1}\), and add up the results. It gives the exact same 93 found using the formula.

It's easy to mix these two terms up, so keep this distinction in mind: a geometric sequence is the list of terms, such as \(3, 6, 12, 24, 48\), while a geometric series is what you get when you add those terms together, giving a single number like \(93\). Understanding the sequence first makes the series much easier to follow.

The formula on this page always works for a finite number of terms, no matter what \(r\) is (as long as \(r \ne 1\)). But something special happens when \(-1 < r < 1\): as \(n\) gets larger and larger, \(r^n\) shrinks toward zero, so the sum approaches a fixed limiting value even as more and more terms are added. That situation, summing infinitely many terms, is covered in the lesson on infinite geometric series.

• Mixing up \(a_1\) (the first term) with the common ratio \(r\) when substituting into the formula.
• Forgetting that the exponent on \(r\) is \(n\), the number of terms being summed, not \(n - 1\).
• Trying to use the standard formula when \(r = 1\); use \(S_n = n \times a_1\) instead.
• Confusing the sum of a geometric series with the sum of an arithmetic series, which has a different pattern of growth between terms.

Related lessons

• Algebra Help