C. Discrete Functions · Sequences and Series

Geometric Series

C2.3 C2.4About 75 minutes

Learning goals

I can develop the formula for the sum of a geometric series.
I can use Sₙ = a(rⁿ − 1)/(r − 1) to find sums, including when r is a fraction or negative.
I can find the number of terms of a geometric series.
I can describe what happens to the partial sums when 0 < r < 1.

Words to know

geometric series
The sum of the terms of a geometric sequence, such as 3 + 6 + 12 + 24.

sum of a geometric series
Sₙ = a(rⁿ − 1)/(r − 1), or equivalently a(1 − rⁿ)/(1 − r), for r ≠ 1.

partial sum
Sₙ, the sum of the first n terms.

The chessboard legend

An old legend tells of a ruler who offered a reward to the inventor of chess. The inventor asked for rice: 1 grain on the first square of the board, 2 on the second, 4 on the third, doubling on each of the 64 squares. The ruler laughed at such a small request, until the counting began.

The total is 1 + 2 + 4 + 8 + … + 2⁶³, a geometric series. Pairing first and last terms (the trick for arithmetic series) does not work here, because the terms are not evenly spaced. We need a new trick. By the end of the lesson you will find the total in one line.

Developing the formula

Start with a number example: S = 3 + 6 + 12 + 24 + 48 (a = 3, r = 2).

Multiply every term by r = 2: 2S = 6 + 12 + 24 + 48 + 96.

Subtract the first line from the second. Almost everything cancels: 2S − S = 96 − 3, so S = 93.

Now do the same with letters. Let Sₙ = a + ar + ar² + … + arⁿ⁻¹.

Multiply by r: rSₙ = ar + ar² + … + arⁿ⁻¹ + arⁿ
Subtract: rSₙ − Sₙ = arⁿ − a
Factor: Sₙ(r − 1) = a(rⁿ − 1)
Divide (r ≠ 1):

Sₙ = a(rⁿ − 1) ÷ (r − 1)

The same formula can be written Sₙ = a(1 − rⁿ) ÷ (1 − r) (multiply the top and bottom by −1). This form avoids negative numbers when 0 < r < 1.

If r = 1, every term equals a, so Sₙ = na.

Multiply the series by r = 2 and line it up under the original. All the middle terms cancel when you subtract, leaving only the last term of 2S and the first term of S.

Worked example

Example 1: A growing series

Find the sum of the first 10 terms of 3 + 6 + 12 + ….

a = 3, r = 2, n = 10.
S₁₀ = 3(2¹⁰ − 1) ÷ (2 − 1) = 3(1024 − 1) ÷ 1 = 3(1023) = 3069.

S₁₀ = 3069

Worked example

Example 2: A shrinking series

Find the sum of the first 8 terms of 64 + 32 + 16 + ….

a = 64, r = 1/2, n = 8. Use Sₙ = a(1 − rⁿ) ÷ (1 − r).
S₈ = 64(1 − (1/2)⁸) ÷ (1 − 1/2) = 64(1 − 1/256) ÷ (1/2)
= 128(255/256) = 127.5
(Check: 64 + 32 + 16 + 8 + 4 + 2 + 1 + 0.5 = 127.5 ✓)

S₈ = 127.5

Worked example

Example 3: A negative ratio

Find the sum of the first 6 terms of 2 − 6 + 18 − ….

a = 2, r = −3, n = 6.
S₆ = 2((−3)⁶ − 1) ÷ (−3 − 1) = 2(729 − 1) ÷ (−4) = 2(728) ÷ (−4) = −364.
Check: 2 − 6 + 18 − 54 + 162 − 486 = −364 ✓. Use brackets for (−3)⁶ on a calculator.

S₆ = −364

Worked example

Example 4: Find n first

Find 5 + 15 + 45 + … + 10 935.

a = 5, r = 3. Find n: 5(3)ⁿ⁻¹ = 10 935 → 3ⁿ⁻¹ = 2187 = 3⁷ → n = 8.
S₈ = 5(3⁸ − 1) ÷ (3 − 1) = 5(6561 − 1) ÷ 2 = 5(6560) ÷ 2 = 16 400.

16 400

Worked example

Example 5: The chessboard

How many grains of rice are on the whole chessboard: 1 + 2 + 4 + … + 2⁶³?

a = 1, r = 2, n = 64.
S₆₄ = 1(2⁶⁴ − 1) ÷ (2 − 1) = 2⁶⁴ − 1.
On a calculator, 2⁶⁴ ≈ 1.8 × 10¹⁹, so the total is about 18 000 000 000 000 000 000 grains, far more rice than the world grows in a year.

2⁶⁴ − 1 ≈ 1.8 × 10¹⁹ grains

Worked example

Example 6: Find the number of terms from the sum

How many terms of 1 + 3 + 9 + … add up to 3280?

a = 1, r = 3: Sₙ = (3ⁿ − 1) ÷ 2 = 3280.
3ⁿ − 1 = 6560 → 3ⁿ = 6561 = 3⁸ → n = 8.

8 terms

The series 1/2 + 1/4 + 1/8 + … fills a 1-by-1 square, half of what is left each time. The partial sums 1/2, 3/4, 7/8, 15/16, … get closer and closer to 1 but never pass it.

When 0 < r < 1

For 1/2 + 1/4 + 1/8 + …, the formula gives Sₙ = (1/2)(1 − (1/2)ⁿ) ÷ (1/2) = 1 − (1/2)ⁿ. As n grows, (1/2)ⁿ shrinks towards 0, so the partial sums approach 1. You met this idea of a limit in Grade 9. When 0 < r < 1 (or −1 < r < 0), adding more and more terms changes the sum less and less. You will use this when a bouncing ball travels a total distance in the problem-solving lessons.

Just for fun

An infinite number of mathematicians walk into a café. The first orders a pizza, the second orders half a pizza, the third a quarter of a pizza, and so on.

The server brings out 2 pizzas and says, "You all need to know your limits."

Hands-on activity

Rice on a chessboard

You need: a chessboard or checkerboard (or an 8 × 8 grid drawn on paper), uncooked rice in a bowl, and a calculator.

Put 1 grain on the first square, 2 on the second, 4 on the third, 8 on the fourth, and keep doubling as long as you have patience (try to get to the 8th square, which needs 128 grains).
Before counting, use the formula to predict the total number of grains on the first 8 squares. Then count all the grains you placed. Did they match?
Use the formula to find the total for the first row of 8 squares, the first two rows (16 squares), and the whole board.
A cup of uncooked long-grain rice holds very roughly 8000 grains (it depends on the rice). About how many cups would the first 16 squares need?

What to look for: first 8 squares: 2⁸ − 1 = 255 grains. 16 squares: 2¹⁶ − 1 = 65 535 grains, about 65 535 ÷ 8000 ≈ 8 cups. The whole board: 2⁶⁴ − 1 ≈ 1.8 × 10¹⁹ grains. (Put the rice back in the bowl to cook later!)

Your turn

Find the sum of the first 7 terms of 2 + 6 + 18 + ….

Find the sum of the first 6 terms of 1 + 5 + 25 + ….

Find the sum of the first 5 terms of 81 + 27 + 9 + ….

Find the sum of the first 8 terms of 1 − 2 + 4 − 8 + ….

Find 4 + 8 + 16 + … + 1024.

Find the sum of the first 12 terms of 7 + 7 + 7 + ….

How many terms of 2 + 4 + 8 + … add up to 510?

More practice: warm-up

Write the formula for the sum of a geometric series.

Find 1 + 2 + 4 + 8 + 16 + 32.

For 3 + 12 + 48 + …, what are a and r?

Find the sum of the first 4 terms of 10 + 30 + 90 + ….

Why can't you use the formula when r = 1?

More practice: on your own

Find the sum of the first 10 terms of 0.5 + 1 + 2 + ….

Find 6 − 12 + 24 − … + 1536.

Find the sum of the first 6 terms of 1000 + 100 + 10 + ….

A geometric series has a = 3 and r = 2, and its sum is 765. How many terms are there?

Find the sum of the first 5 terms of a geometric series with t₂ = 6 and t₃ = 18.

Challenge

Show that 1 + 2 + 4 + … + 2ⁿ⁻¹ is always one less than the next term, 2ⁿ. What does this mean for the chessboard?

The sum of the first 3 terms of a geometric series is 26 and the first term is 2. Find the possible values of r.

For 6 + 3 + 1.5 + …, find S₁₀ and explain what number the partial sums approach.

Check your understanding

Quiz

What is the sum of the first 5 terms of 1 + 3 + 9 + …?

In the derivation of the formula, why do we multiply Sₙ by r?

Find the sum of the first 6 terms of 32 + 16 + 8 + ….

Find the sum of the first 4 terms of 5 − 10 + 20 − ….

A geometric series has r = 1, a = 9 and n = 20. What is its sum?

The partial sums of 1/2 + 1/4 + 1/8 + … approach:

For parents and teachers

Big idea: Expectation C2.3 asks students to develop the formula for the sum of a geometric series through investigation, connecting the steps of a number example to the algebra, and to use it to find the sum of a given number of terms. The "multiply by r and subtract" trick is the heart of the lesson. The same formula drives the annuity calculations in the Financial Applications unit.

How to help: Ask your teen to show you the multiply-and-subtract trick with 1 + 2 + 4 + 8 + 16. Then ask them to estimate the chessboard total before calculating it.

Common difficulties:

Calculator errors with negative ratios: (−3)⁶ needs brackets.
Mixing up n and the last term: find n first (Example 4).
Using the r − 1 form with a fraction and then mishandling the negative numbers: the 1 − r form is easier when r < 1.
Trying to use the formula when r = 1.