Geometric Series

A geometric series is the sum of the terms of a geometric sequence, where each term is the one before it times a common ratio r. The sum of the first n terms is S_n = \frac{a_1(1 - r^n)}{1 - r}, where a_1 is the first term and r \ne 1.

Before this lesson: You should know Geometric Sequences, Sigma Notation and Arithmetic Series.

Why the Formula Works

In an arithmetic series the terms go up by adding the same number. In a geometric series they grow (or shrink) by multiplying by the same number. So the pairing trick from arithmetic series does not work here. A different trick does: multiply the whole sum by r and subtract.

1. Write the sum: S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n - 1}
2. Multiply every term by r: rS_n = a_1 r + a_1 r^2 + \cdots + a_1 r^{n - 1} + a_1 r^n
3. Subtract the second line from the first. Every middle term cancels: S_n - rS_n = a_1 - a_1 r^n
4. Factor out S_n on the left and a_1 on the right (see factoring out the GCF): S_n(1 - r) = a_1(1 - r^n)
5. Divide both sides by 1 - r: S_n = \frac{a_1(1 - r^n)}{1 - r}

The Geometric Series Formula

S_n = \frac{a_1(1 - r^n)}{1 - r}, \quad r \ne 1

If r = 1, every term is the same, so the sum is just n \cdot a_1.

Worked Examples

Example 1: Sum of a growing series

Find the sum of the first 6 terms of 3, 6, 12, 24, \ldots

r = \frac{6}{3} = 2

Identify the parts: a_1 = 3, r = 2, n = 6

S_6 = \frac{3(1 - 2^6)}{1 - 2}

S_6 = \frac{3(1 - 64)}{1 - 2}

S_6 = \frac{3(-63)}{-1}

S_6 = \frac{-189}{-1}

S_6 = 189

Answer: 189. Check: 3 + 6 + 12 + 24 + 48 + 96 = 189.

Example 2: A ratio between 0 and 1

Find the sum of the first 5 terms of 80, 40, 20, \ldots

r = \frac{40}{80} = \frac{1}{2}

S_5 = \frac{80(1 - (\frac{1}{2})^5)}{1 - \frac{1}{2}}

S_5 = \frac{80(1 - \frac{1}{32})}{1 - \frac{1}{2}}

S_5 = \frac{80 \cdot \frac{31}{32}}{\frac{1}{2}}

S_5 = \frac{\frac{155}{2}}{\frac{1}{2}}

S_5 = \frac{155}{2} \cdot 2 = 155

Answer: 155. Check: 80 + 40 + 20 + 10 + 5 = 155.

Example 3: A negative ratio in sigma notation

Evaluate \sum_{n=1}^{6} 2(-3)^{n - 1}.

2(-3)^0 = 2

r = -3

n = 6

S_6 = \frac{2(1 - (-3)^6)}{1 - (-3)}

(-3)^6 = 729

2(1 - 729) = 2(-728) = -1456

1 - (-3) = 4

\frac{-1456}{4} = -364

Answer: −364. Check: 2 - 6 + 18 - 54 + 162 - 486 = -364.

Example 4: Find the number of terms

How many terms of 4 + 12 + 36 + \cdots add up to 484?

r = \frac{12}{4} = 3

484 = \frac{4(1 - 3^n)}{1 - 3}

484 = \frac{4(1 - 3^n)}{-2}

484 = -2(1 - 3^n)

-242 = 1 - 3^n

-243 = -3^n

243 = 3^n

3^5 = 3^n

n = 5

Answer: 5 terms. Check: 4 + 12 + 36 + 108 + 324 = 484.

Example 5: Word problem with yearly raises

A job pays $40,000 the first year, and the pay goes up 5% each year. What is the total pay for the first 4 years?

A 5% raise multiplies the pay by 1.05 each year, so a_1 = 40{,}000, r = 1.05 and n = 4

S_4 = \frac{40{,}000(1 - 1.05^4)}{1 - 1.05}

1.05 \cdot 1.05 = 1.1025

1.1025 \cdot 1.1025 = 1.21550625

1 - 1.21550625 = -0.21550625

1 - 1.05 = -0.05

\frac{-0.21550625}{-0.05} = 4.310125

40{,}000 \cdot 4.310125 = 172{,}405

Answer: $172,405. Check: 40{,}000 + 42{,}000 + 44{,}100 + 46{,}305 = 172{,}405. Compare this with the arithmetic raises in arithmetic series, where the pay grows by the same dollar amount each year.

Common Mistakes

Related Methods

An equivalent form of the formula

Multiplying the top and bottom by −1 gives S_n = \frac{a_1(r^n - 1)}{r - 1}. This form avoids negative signs when r \gt 1. Both forms always give the same answer.

Using the last term

If you know the last term a_n, you can also use S_n = \frac{a_1 - r a_n}{1 - r}. For Example 1, \frac{3 - 2 \cdot 96}{1 - 2} = \frac{-189}{-1} = 189.

When n gets very large

If r is between −1 and 1, the power r^n gets closer and closer to 0 as n grows. That leads to a formula for adding infinitely many terms, which is the next lesson.

Practice Problems

1. Find the sum of the first 5 terms of 1, 2, 4, \ldots
2. Find the sum of the first 6 terms of 5, 15, 45, \ldots
3. Find the sum of the first 4 terms of 64, 16, 4, \ldots
4. Evaluate \sum_{k=1}^{5} 3(2)^{k - 1}.
5. Evaluate \sum_{n=1}^{4} 10(-2)^{n - 1}.
6. Find the sum of the first 8 terms of 1, -1, 1, -1, \ldots
7. Find 2 + 6 + 18 + \cdots + 486.
8. A geometric series has a_1 = 7 and r = 2. How many terms add up to 217?
9. A geometric series has 4 terms, a ratio of 3 and a sum of 120. Find the first term.
10. A small shop sells $2,000 of goods in its first month, and sales grow 10% each month. What are its total sales for the first 3 months?

r = \frac{2}{1} = 2

S_5 = \frac{1(1 - 2^5)}{1 - 2}

S_5 = \frac{1 - 32}{-1}

S_5 = \frac{-31}{-1}

S_5 = 31

r = \frac{15}{5} = 3

S_6 = \frac{5(1 - 3^6)}{1 - 3}

S_6 = \frac{5(1 - 729)}{-2}

S_6 = \frac{5(-728)}{-2}

S_6 = \frac{-3640}{-2}

S_6 = 1820

r = \frac{16}{64} = \frac{1}{4}

S_4 = \frac{64(1 - (\frac{1}{4})^4)}{1 - \frac{1}{4}}

S_4 = \frac{64(1 - \frac{1}{256})}{\frac{3}{4}}

S_4 = \frac{64 \cdot \frac{255}{256}}{\frac{3}{4}}

S_4 = \frac{\frac{255}{4}}{\frac{3}{4}}

S_4 = \frac{255}{4} \cdot \frac{4}{3} = 85

S_5 = \frac{3(1 - 2^5)}{1 - 2}

S_5 = \frac{3(1 - 32)}{-1}

S_5 = \frac{3(-31)}{-1}

S_5 = \frac{-93}{-1}

S_5 = 93

S_4 = \frac{10(1 - (-2)^4)}{1 - (-2)}

(-2)^4 = 16

10(1 - 16) = 10(-15) = -150

1 - (-2) = 3

\frac{-150}{3} = -50

r = \frac{-1}{1} = -1

S_8 = \frac{1(1 - (-1)^8)}{1 - (-1)}

(-1)^8 = 1

S_8 = \frac{1 - 1}{2} = \frac{0}{2} = 0

r = \frac{6}{2} = 3

2 \cdot 3^{n - 1} = 486

3^{n - 1} = 243

3^{n - 1} = 3^5

n - 1 = 5

n = 6

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

\frac{2(-728)}{-2} = \frac{-1456}{-2} = 728

217 = \frac{7(1 - 2^n)}{1 - 2}

217 = \frac{7(1 - 2^n)}{-1}

217 = -7(1 - 2^n)

-31 = 1 - 2^n

-32 = -2^n

32 = 2^n

2^5 = 2^n

n = 5

120 = \frac{a_1(1 - 3^4)}{1 - 3}

120 = \frac{a_1(1 - 81)}{-2}

120 = \frac{-80a_1}{-2}

120 = 40a_1

a_1 = 3

S_3 = \frac{2000(1 - 1.1^3)}{1 - 1.1}

1.1^3 = 1.1 \cdot 1.1 \cdot 1.1 = 1.331

1 - 1.331 = -0.331

1 - 1.1 = -0.1

\frac{-0.331}{-0.1} = 3.31

2000 \cdot 3.31 = 6620

FAQ

What is the formula for the sum of a geometric series?

The sum of the first n terms is S_n = \frac{a_1(1 - r^n)}{1 - r}, where a_1 is the first term and r is the common ratio. It works for any ratio except r = 1.

How do you find the common ratio of a geometric series?

Divide any term by the term just before it. For 5 + 15 + 45 + \cdots, the ratio is \frac{15}{5} = 3.

What is the difference between a geometric series and an arithmetic series?

In an arithmetic series you add the same number to get each new term. In a geometric series you multiply by the same number. They use different sum formulas.

What happens if the common ratio is 1?

Every term equals the first term, so the formula would divide by zero. Instead, multiply: the sum is n \cdot a_1.

Summary

• A geometric series adds the terms of a geometric sequence.
• Find the ratio by dividing a term by the one before it.
• S_n = \frac{a_1(1 - r^n)}{1 - r} for r \ne 1.
• Watch the signs when r is negative: even powers are positive.
• To find n, set up the formula and solve the exponential equation.

Previous lesson: Arithmetic Series

Next lesson: Infinite Geometric Series

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