Geometric Series

A geometric series is the sum of a geometric sequence, like $3 + 6 + 12 + 24$. Because the terms multiply, these sums can grow astonishingly fast, and they’re the math behind savings plans, loans, and the classic grains-of-rice puzzle.

Key ideas

The formula

For a geometric series with first term $a$ and common ratio $r \ne 1$, the sum of the first $n$ terms is:

$$S_n = \frac{a(r^n - 1)}{r - 1}$$

The same formula can be written $S_n = \dfrac{a(1 - r^n)}{1 - r}$, which is handier when $0 \lt r \lt 1$ because it avoids negative numbers. (If $r = 1$, every term is $a$, so $S_n = na$.)

Why it works

Write $S_n$, then multiply it by $r$:

$$\begin{aligned} S_n &= a + ar + ar^2 + \dots + ar^{n - 1} \\ rS_n &= \phantom{a + {}} ar + ar^2 + \dots + ar^{n - 1} + ar^n \end{aligned}$$

Subtract the first line from the second. Everything in the middle cancels:

$$rS_n - S_n = ar^n - a \quad\Rightarrow\quad S_n(r - 1) = a(r^n - 1) \quad\Rightarrow\quad S_n = \frac{a(r^n - 1)}{r - 1}$$

Finding n first

If you’re given the last term instead of $n$, use $t_n = ar^{n - 1}$ to find $n$, then use the sum formula.

Worked examples

Example 1: A growing series

Find the sum of the first $8$ terms of $3 + 6 + 12 + \dots$

Solution. $a = 3$, $r = 2$, $n = 8$:

$$S_8 = \frac{3(2^8 - 1)}{2 - 1} = 3(255) = 765$$

Example 2: A shrinking series

Find the sum of the first $7$ terms of $128 + 64 + 32 + \dots$

Solution. $a = 128$, $r = \tfrac{1}{2}$, $n = 7$. Use the second form of the formula:

$$S_7 = \frac{128\left(1 - \left(\frac{1}{2}\right)^7\right)}{1 - \frac{1}{2}} = \frac{128\left(1 - \frac{1}{128}\right)}{\frac{1}{2}} = 256\left(\frac{127}{128}\right) = 254$$

Check by adding: $128 + 64 + 32 + 16 + 8 + 4 + 2 = 254$. ✓

Example 3: A negative ratio

Find the sum of the first $6$ terms of $2 - 6 + 18 - \dots$

Solution. $a = 2$, $r = -3$, $n = 6$:

$$S_6 = \frac{2\big((-3)^6 - 1\big)}{-3 - 1} = \frac{2(729 - 1)}{-4} = \frac{1456}{-4} = -364$$

Example 4: Finding n first

Find the sum $1 + 3 + 9 + \dots + 6561$.

Solution. $a = 1$, $r = 3$. Find $n$:

$$3^{n - 1} = 6561 = 3^8 \quad\Rightarrow\quad n = 9$$ $$S_9 = \frac{1(3^9 - 1)}{3 - 1} = \frac{19\,683 - 1}{2} = 9841$$

Common mistakes

Using $r^{n - 1}$ in the sum formula. The term formula has $r^{n - 1}$; the sum formula has $r^n$.

Mishandling a negative ratio. Keep $r$ in brackets: $(-3)^6 = 729$, and the denominator is $-3 - 1 = -4$.

Using the formula when $r = 1$. The denominator would be $0$. If every term is the same, just multiply: $S_n = na$.

Miscounting $n$. In Example 4, the last term is $3^8$, but it’s the $9$th term, because the first term is $3^0$.

Rounding the ratio. If $r = \tfrac{2}{3}$, keep it as a fraction rather than $0.67$, or your answer will drift.

Practice

1. (Warm-up) Find the sum of the first $5$ terms of $1 + 2 + 4 + \dots$

Solution

$$S_5 = \frac{1(2^5 - 1)}{2 - 1} = 31$$

Check: $1 + 2 + 4 + 8 + 16 = 31$. ✓

2. (Warm-up) Find the sum of the first $6$ terms of $5 + 15 + 45 + \dots$

Solution

$$S_6 = \frac{5(3^6 - 1)}{3 - 1} = \frac{5(728)}{2} = 1820$$

3. (Warm-up) Is $4 + 8 + 16 + \dots$ an arithmetic or a geometric series? Give $a$ and $r$ or $d$.

Solution

Geometric, with $a = 4$ and $r = 2$. (The differences $4, 8$ aren’t constant, but the ratios are.)

4. (Core) Find the sum of the first $8$ terms of $64 + 32 + 16 + \dots$

Solution

$$S_8 = \frac{64\left(1 - \left(\frac{1}{2}\right)^8\right)}{\frac{1}{2}} = 128\left(\frac{255}{256}\right) = 127.5$$

5. (Core) Find the sum of the first $7$ terms of $1 - 2 + 4 - 8 + \dots$

Solution

$$S_7 = \frac{1\big((-2)^7 - 1\big)}{-2 - 1} = \frac{-128 - 1}{-3} = \frac{-129}{-3} = 43$$

6. (Core) Find the sum $2 + 6 + 18 + \dots + 4374$.

Solution

$2(3)^{n - 1} = 4374$ gives $3^{n - 1} = 2187 = 3^7$, so $n = 8$.

$$S_8 = \frac{2(3^8 - 1)}{3 - 1} = 6561 - 1 = 6560$$

7. (Core) A ball is dropped from $10$ m and rebounds to $80\%$ of its previous height each time. How far has it travelled up and down when it hits the ground for the $5$th time?

Solution

It falls $10$ m first. Then it rises and falls the same distance after each of the first four bounces: $8$, $6.4$, $5.12$, and $4.096$ m.

The rebound heights form a geometric series with $a = 8$, $r = 0.8$, $n = 4$:

$$S_4 = \frac{8(1 - 0.8^4)}{1 - 0.8} = \frac{8(0.5904)}{0.2} = 23.616$$

Total distance $= 10 + 2(23.616) = 57.232$ m, or about $57.2$ m.

8. (Challenge) A legend says a king agreed to put $1$ grain of rice on the first square of a chessboard, $2$ on the second, $4$ on the third, and so on, doubling for all $64$ squares. How many grains is that in total?

Solution

$$S_{64} = \frac{1(2^{64} - 1)}{2 - 1} = 2^{64} - 1 = 18\,446\,744\,073\,709\,551\,615$$

That’s about $1.8 \times 10^{19}$ grains, far more rice than has ever been grown on Earth.

9. (Challenge) Use the “multiply by $r$ and subtract” method to find $1 + 5 + 25 + 125 + 625$ without the formula.

Solution

$$\begin{aligned} S &= 1 + 5 + 25 + 125 + 625 \\ 5S &= \phantom{1 + {}} 5 + 25 + 125 + 625 + 3125 \end{aligned}$$

Subtracting, $5S - S = 3125 - 1$, so $4S = 3124$ and $S = 781$.

Check by adding: $1 + 5 + 25 + 125 + 625 = 781$. ✓