Sum of the First n Terms of Geometric Sequences

Finding the sum of the first $n$ terms of a geometric sequence is an important problem in the study of sequences. Unlike arithmetic sequences, the summation formula of a geometric sequence needs to be discussed in separate cases.

Grains of Rice on the Chessboard

Let us return to the story from the opening: on the 64 squares of a chessboard, put 1 grain of rice on the first square, 2 grains on the second, 4 grains on the third… each square has twice as many as the previous one.

This forms a geometric sequence: $1, 2, 4, 8, 16, \ldots, 2^{63}$

How many grains of rice are needed in total? That is, find:

$S_{64} = 1 + 2 + 4 + 8 + \cdots + 2^{63}$

How do we compute this sum? Let us derive the summation formula of a geometric sequence.

Derivation of the Formula

Let $S_n$ be the sum of the first $n$ terms of the geometric sequence $\{a_n\}$, with first term $a_1$ and common ratio $q$.

Case 1: $q = 1$

When $q = 1$, the sequence is constant: $a_1, a_1, a_1, \ldots$

In this case:

Case 2: $q \neq 1$

When $q \neq 1$, use the method of “multiplying by the common ratio and subtracting”:

$S_n = a_1 + a_1 q + a_1 q^2 + \cdots + a_1 q^{n-1}$

Multiply both sides by $q$:

$qS_n = a_1 q + a_1 q^2 + a_1 q^3 + \cdots + a_1 q^n$

Subtract the two equations (the second minus the first):

$$\begin{aligned} qS_n - S_n &= a_1 q^n - a_1 \\ S_n(q - 1) &= a_1(q^n - 1) \\ S_n &= \frac{a_1(q^n - 1)}{q - 1} \end{aligned}$$

It can also be written as:

$S_n = \frac{a_1(1 - q^n)}{1 - q}$

The Answer to the Chessboard Problem

Now we can compute the total number of grains of rice on the chessboard:

$$\begin{aligned} S_{64} &= \frac{1 \times (2^{64} - 1)}{2 - 1} \\ &= 2^{64} - 1 \\ &\approx 1.84 \times 10^{19} \end{aligned}$$

This number is about 18,446,744,073,709,551,615 grains!

If each grain of rice weighs 0.02 grams, the total weight is about $3.69 \times 10^{14}$ kilograms, equivalent to 369 billion tons—far exceeding the world’s annual grain production!

Worked Examples

Example 1: Using the Summation Formula

Find the sum of the first 6 terms of the geometric sequence $2, 6, 18, 54, \ldots$.

Solution:

Given $a_1 = 2$, $q = 3$, $n = 6$.

Using the formula (for $q > 1$, use the second form):

$S_6 = \frac{2 \times (3^6 - 1)}{3 - 1} = \frac{2 \times (729 - 1)}{2} = \frac{2 \times 728}{2} = 728$

Example 2: Finding the Number of Terms from the Sum

In the geometric sequence $\{a_n\}$, $a_1 = 1$ and $q = 2$. If the sum of the first $n$ terms is $S_n = 127$, find $n$.

Solution:

$S_n = \frac{1 \times (2^n - 1)}{2 - 1} = 2^n - 1 = 127$

$2^n = 128 = 2^7$

$n = 7$

Example 3: The Sum of an Infinitely Decreasing Geometric Sequence

When $|q| < 1$, as $n$ increases, $q^n$ tends to 0, and the sum of the first $n$ terms approaches a limiting value:

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{a_1(1 - q^n)}{1 - q} = \frac{a_1}{1 - q}$

Example: Find the sum of $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$.

This is an infinite geometric sequence with first term $a_1 = \frac{1}{2}$ and common ratio $q = \frac{1}{2}$.

$S = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$

Practice Problems

Exercise 1

Find the sum of the first 8 terms of the geometric sequence $1, 3, 9, 27, \ldots$.

Exercise 2

In the geometric sequence $\{a_n\}$, $a_1 = 64$ and $q = \frac{1}{2}$. Find the sum of the first 6 terms.

Exercise 3

Find the sum of the infinite geometric sequence $1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots$.

Exercise 4

Adapted from a postgraduate entrance examination problem

The sum of the first $n$ terms of the geometric sequence $\{a_n\}$ is $S_n$. Given $S_3 = 7$ and $S_6 = 63$, find the common ratio $q$.

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