Year 11 Extended · Unit 3 Sequences, series, formulae and proportion · Lesson 4 of 7

What cancels when a sum is multiplied by its ratio?

A geometric sequence grows too fast to add term by term for long. This lesson finds a formula for its sum by a cancelling trick, and uses it forwards and backwards.

Number and AlgebraNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 10: find the common ratio of 5, 15, 45, …
    Answer3.
  2. From Year 9: evaluate 27.
    Answer128.
  3. From Year 11 Standard: find the 4th term of 2, 6, 18, …
    Answer54.
Words used in this lesson
geometric series
the sum of the terms of a geometric sequence. For example: 1 + 3 + 9.
common ratio
the number each term is multiplied by. For example: r = 2.
cancel
two equal parts of opposite sign add to nothing. For example: 7 − 7.
shifted
written one place along. For example: under the next box.
formula
a rule written with letters that gives a sum from a, r and n. For example: Sn.

How many people can one rumour reach?

In a village in Kerala, only Anil knows a piece of news at 9 o'clock. In each hour after that, every person who learned it in the previous hour tells exactly 2 people who have not heard it, and then stops. So the news spreads in waves.

Predict first: by 3 o'clock in the afternoon, six hours later, do you expect tens, hundreds or thousands of people to know? Then use the figure: it writes a sum S and the same sum times its ratio one place along. Which terms line up?

2
3
4
80the sum S
160rS minus S

First term 2, ratio 3, 4 terms: S = 80 and 3S = 240. The middle terms cancel, so 3S − S = 162 − 2 = 160, and S = 160 ÷ 2 = 80.

Show me the steps
  1. First term 2, ratio 3, 4 terms: S = 2 + 6 + 18 + 54 = 80 and 3S = 240. The middle terms cancel, so 3S − S = 162 − 2 = 160, and S = 160 ÷ 2 = 80.
  2. First term 1, ratio 2, 5 terms: S = 1 + 2 + 4 + 8 + 16 = 31 and 2S = 62. Then 2S − S = 32 − 1 = 31, and dividing by r − 1 = 1 leaves S = 31.
  3. First term 4, ratio 4, 3 terms: S = 4 + 16 + 64 = 84 and 4S = 336. Then 4S − S = 256 − 4 = 252, and S = 252 ÷ 3 = 84.
  4. Any first term a, ratio r and n terms: rS − S = arn − a, so S = a(rn − 1) ÷ (r − 1).

Multiply by the ratio, then subtract. Write S = a + ar + ar2 + … + arn−1. Then rS is the same terms shifted one place: ar + ar2 + … + arn. Subtract, and every term cancels except the first of S and the last of rS:

rS − S = arn − a, so Sn = a(rn − 1) ÷ (r − 1).

The formula needs r not equal to 1. When r = 1 every term is a and the sum is na.

Two common slips. Using rn−1 in the formula: that is the nth term, and the sum has rn. And leaving out the −1: for 1 + 3 + 9 the formula gives (27 − 1) ÷ 2 = 13, which is right, but 27 ÷ 2 is not.

Worked examples

1. Find 2 + 6 + 18 + 54 + 162 + 486.

  1. The first term is 2, the ratio is 3 and there are 6 terms.
  2. S = 2 × (36 − 1) ÷ (3 − 1) = 2 × 728 ÷ 2 = 728.

2. Find the sum of the first 6 terms of 7, 14, 28, …

  1. The first term is 7 and the ratio is 2.
  2. S = 7 × (26 − 1) ÷ (2 − 1) = 7 × 63 = 441.

3. Find 64 + 32 + 16 + 8 + 4 + 2.

  1. The ratio is 1 ÷ 2, so r is less than 1. S = 64 × (1 − (1 ÷ 2)6) ÷ (1 − 1 ÷ 2).
  2. (1 ÷ 2)6 = 1 ÷ 64, so S = 64 × (63 ÷ 64) ÷ (1 ÷ 2) = 63 × 2 = 126.

Your go. The first step is given. Find 3 + 12 + 48 + 192 + 768. The ratio is 4 and there are 5 terms.

  1. S = 3 × (45 − 1) ÷ (4 − 1).
  2. Work out 45 and finish.
Check the answer

45 = 1024, so S = 3 × 1023 ÷ 3 = 1023.

Check before you go on

1. Which calculation gives the sum 5 + 15 + 45 + … + 3645?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. A chain message is sent to 5 people. Each of them sends it to 5 new people, and so on. Counting the 5 people in the first round, how many messages have been sent after 4 rounds?

3. Find the sum of the first 5 terms of 243, 81, 27, …

Level 2 · Building

4. How many terms of 2 + 4 + 8 + … are needed for a sum of 510?

5. The first term of a geometric sequence is 6 and the sum of its first 3 terms is 78. Find the positive common ratio.

Level 3 · Stretch

6. The sum of the first 4 terms of a geometric sequence with a positive ratio is 15 times its first term. Find the common ratio.

7. The first term of a geometric sequence is 3 and the common ratio is 2. What is the smallest number of terms whose sum is greater than 1000?

Something to think about

The formula divides by r − 1. What goes wrong when the ratio is 1, and what is the sum then?

One way to think about it

The division by zero is the problem: 1 − 1 = 0. But when r = 1 every term is a, so the sum is just n × a. The formula is for r not equal to 1, and r = 1 is a separate case.

Exit ticket

8. Let S = 7 + 21 + 63 + … + 7 × 319. Which expression is equal to 3S − S?