Geometric Series: The Sum Formula and Loan Payments

Adding a geometric sequence

A geometric sequence multiplies by the same ratio each step. Adding its terms makes a geometric series.

$$3 + 6 + 12 + 24 + 48 + 96$$

Here the first term is $a = 3$, the ratio is $r = 2$, and there are $n = 6$ terms. Adding them one by one gives $189$. That is fine for six terms and hopeless for sixty.

A legend makes the point. A king offers one grain of rice for the first square of a chessboard, two for the second, four for the third, doubling across all $64$ squares. The total is

$$1 + 2 + 4 + \cdots + 2^{63}$$

which no one is going to add term by term.

Why multiplying by r and subtracting works

Write the sum of $n$ terms with first term $a$ and ratio $r$:

$$S = a + ar + ar^2 + \cdots + ar^{n-1}$$

Multiply every term by $r$. Each term becomes the next one:

$$rS = ar + ar^2 + \cdots + ar^{n-1} + ar^n$$

The two lines share every term from $ar$ to $ar^{n-1}$. Subtract, and all of those cancel. Only the first term of $S$ and the last term of $rS$ survive:

$$S - rS = a - ar^n$$

Factor both sides and divide by $1 - r$:

$$S(1 - r) = a(1 - r^n) \qquad\Longrightarrow\qquad S = \frac{a(1 - r^n)}{1 - r}, \quad r \ne 1$$

The whole trick is that multiplying by $r$ shifts the series one place. A shifted copy lines up with the original everywhere except at the two ends.

The condition $r \ne 1$ is not a technicality: the last step divides by $1 - r$. When $r = 1$ every term is $a$, and the sum is $na$ directly.

The chessboard

For the rice, $a = 1$, $r = 2$ and $n = 64$:

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

That is over eighteen quintillion grains — hundreds of years of the whole world’s rice harvest. Doubling looks harmless for the first few squares and then overwhelms everything.

Saving with interest

Deposit $200$ dollars at the end of each month into an account that earns $0.5\%$ a month. After $36$ months, how much is there?

The last deposit has earned nothing. The one before it has grown for one month, to $200(1.005)$. The first has grown for $35$ months. Listed from the last deposit back to the first, the account holds

$$200 + 200(1.005) + 200(1.005)^2 + \cdots + 200(1.005)^{35}$$

That is a geometric series with $a = 200$, $r = 1.005$ and $n = 36$:

$$S = \frac{200(1 - 1.005^{36})}{1 - 1.005} = \frac{200(1.005^{36} - 1)}{0.005} \approx 7{,}867.22$$

The deposits total $7{,}200$ dollars, and interest has added $667.22$.

Loan payments

A car loan works the same way from the other side. Borrow $20{,}000$ dollars at $0.5\%$ a month and repay it in $60$ equal monthly payments $P$.

Picture the lender’s side. The $20{,}000$ grows for $60$ months to $20{,}000(1.005)^{60}$. Each payment also grows from the month it arrives, exactly like a savings deposit. The loan is paid off when the grown payments match the grown loan:

$$P \cdot \frac{1.005^{60} - 1}{0.005} = 20{,}000(1.005)^{60}$$

The left side is the savings series from the last section, with $P$ in place of $200$. Solving for $P$:

$$P = \frac{20{,}000 \times 0.005 \times 1.005^{60}}{1.005^{60} - 1} \approx 386.66$$

The payment is $386.66$ dollars a month. Over $60$ months that is $23{,}199.36$ dollars, so the loan costs about $3{,}199$ dollars in interest.

The same equation, with a larger loan and more months, sets a mortgage payment.

Worked examples

Common mistakes

Practice problems