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Master: Multiplying by the same amount, geometric sequences

Patterns and sequences · step 4 of 5

A question to start

An old story tells of a king 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 every square up to the 64th. The king laughed at so small a wish. How much rice was it really? (Answer below: more rice than the whole world has ever grown.)

The idea

What makes a sequence geometric

A sequence is geometric when you multiply by the same number each time to get the next term. That number is the common ratio, written \(r\).

  • 1, 2, 4, 8, 16, ... has \(r = 2\).
  • 3, 15, 75, 375, ... has \(r = 5\).
  • 80, 40, 20, 10, 5, ... has \(r = \tfrac12\). Multiplying by a number between 0 and 1 makes the terms shrink.
  • 1, −3, 9, −27, ... has \(r = -3\). A negative ratio makes the signs flip back and forth.

To test whether a sequence is geometric, divide each term by the one before it. If every answer is the same, it is geometric.

The formula for any term

As before, reaching position \(n\) takes \(n-1\) steps, but now each step multiplies by \(r\):

\[a_n = a_1 \, r^{\,n-1}\]

\(r^{\,n-1}\) means \(r\) multiplied by itself \(n-1\) times. Compare it with the arithmetic formula \(a_1 + (n-1)d\): adding \(d\) again and again becomes multiplying by \(r\) again and again.

Why geometric sequences get huge (or tiny) so fast

Adding 2 each time, 30 steps take you from 1 to 61. Doubling each time, 30 steps take you from 1 to \(2^{29} = 536\,870\,912\). That is the difference between linear growth (arithmetic) and exponential growth (geometric). Exponential growth always wins in the end, however slowly it starts.

Adding up a geometric sequence

Call the sum \(S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}\). Multiply every term by \(r\), then subtract:

\[ \begin{aligned} S_n &= a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}\\ r\,S_n &= \phantom{a_1 + {}} a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1} + a_1 r^{n}\\ S_n - r\,S_n &= a_1 - a_1 r^{n} \end{aligned} \]

Almost everything cancels. Taking out the common factors, \(S_n(1-r) = a_1(1-r^n)\), so

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

Sums that go on forever

Walk halfway to a wall, then half of what is left, then half again. You walk \(\tfrac12 + \tfrac14 + \tfrac18 + \cdots\) of the distance. You never take a last step, yet you never pass the wall either.

When \(r\) is between −1 and 1, \(r^n\) gets closer and closer to 0 as \(n\) grows. Then the sum formula settles down to

\[S_\infty = \frac{a_1}{1 - r} \qquad (-1 < r < 1)\]

For the walk, \(a_1 = \tfrac12\) and \(r = \tfrac12\), so \(S_\infty = \dfrac{1/2}{1/2} = 1\): the whole distance, exactly. This is also why \(0.999\ldots\) (nines forever) equals 1: it is \(\tfrac{9}{10} + \tfrac{9}{100} + \cdots\), with \(a_1 = \tfrac{9}{10}\) and \(r = \tfrac{1}{10}\), so the sum is \(\dfrac{9/10}{9/10} = 1\).

Worked examples

1. The rice on the chessboard. \(a_1 = 1\), \(r = 2\), \(n = 64\):

\[S_{64} = \frac{1\times(1 - 2^{64})}{1 - 2} = 2^{64} - 1 = 18\,446\,744\,073\,709\,551\,615 \text{ grains.}\]

At roughly 50 000 grains in a kilogram, that is about 370 billion tonnes, hundreds of years of the whole world's rice harvest.

2. The power source on a Mars rover. The Curiosity and Perseverance rovers are powered by the heat of plutonium-238, which has a half-life of about 88 years: every 88 years, half of what is left decays. The amounts after 0, 88, 176, 264 years form a geometric sequence with \(r = \tfrac12\):

\[100\%,\ 50\%,\ 25\%,\ 12.5\%,\ \ldots \qquad a_n = 100\% \times \left(\tfrac12\right)^{n-1}\]

The rovers' power drops much faster than that in practice, because the parts that turn heat into electricity also wear out, but the fuel itself follows this sequence exactly.

3. Finding \(r\) from two terms. A geometric sequence has \(a_2 = 12\) and \(a_5 = 96\). From \(a_2\) to \(a_5\) is 3 steps, so \(r^3 = \dfrac{96}{12} = 8\) and \(r = 2\). Then \(a_1 = \dfrac{12}{2} = 6\): the sequence is 6, 12, 24, 48, 96, ...

Try it

  1. Is 2, 6, 18, 54 geometric? What is \(r\)?
  2. Find \(a_{10}\) for 3, 6, 12, 24, ...
  3. Add \(1 + 3 + 9 + \cdots + 3^{7}\) (8 terms).
  4. A ball is dropped from 2 m and each bounce reaches \(\tfrac34\) of the height before. How far does it travel up and down in total, if it bounces forever?
  5. Write \(0.333\ldots\) as a fraction using the infinite sum formula.
Answers
  1. Yes: \(6 \div 2 = 18 \div 6 = 54 \div 18 = 3\), so \(r = 3\).
  2. \(a_{10} = 3 \times 2^{9} = 3 \times 512 = 1536\).
  3. \(S_8 = \dfrac{1 - 3^8}{1 - 3} = \dfrac{1 - 6561}{-2} = 3280\).
  4. Down 2 m first. After that, each bounce goes up and comes down the same height: the heights are \(1.5, 1.125, \ldots\) with \(a_1 = 1.5\), \(r = \tfrac34\), summing to \(\dfrac{1.5}{1/4} = 6\) m. Up and down doubles that: 12 m. Total \(2 + 12 = 14\) m.
  5. \(0.3 + 0.03 + 0.003 + \cdots\) has \(a_1 = \tfrac{3}{10}\), \(r = \tfrac{1}{10}\): \(\dfrac{3/10}{9/10} = \tfrac13\).

Common mistake

Using the infinite sum when \(r\) is too big. \(1 + 2 + 4 + 8 + \cdots\) has no total: it grows without limit. The formula \(\dfrac{a_1}{1-r}\) would give \(\dfrac{1}{1-2} = -1\), which is nonsense. Only use it when \(r\) is strictly between −1 and 1.

Where it leads