Derive the finite geometric series formula and use it for payments and totals

Learning goal: Derive the formula for the sum of a finite geometric series when the common ratio is not 1, explain why the derivation needs that condition, and use the formula to solve problems, including calculating the level payment on a loan such as a mortgage.

Scope note. A.SSE.4 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.

What a finite geometric series is. A geometric sequence starts at a first term a and multiplies by the same common ratio r at each step: a, ar, ar², …, ar^(n−1). A finite geometric series is the sum of the first n of those terms: S = a + ar + ar² + … + ar^(n−1). The sequence 3, 6, 12, 24, 48, 96, 192, 384 has a = 3, r = 2 and n = 8, and adding directly gives 765. Direct adding is fine for eight terms and hopeless for 360, which is how many monthly payments a 30-year loan has. The formula exists to replace the adding.

Deriving the formula. Write the series and, underneath it, the series multiplied by r:

S = a + ar + ar² + … + ar^(n−1)

rS = ar + ar² + … + ar^(n−1) + ar^n

Every term of the first line except the first, a, appears in the second line, and every term of the second line except the last, ar^n, appears in the first. Subtract the second from the first: S − rS = a − ar^n, because all the shared terms cancel in pairs. Factor both sides: S(1 − r) = a(1 − r^n). Now divide by 1 − r to isolate S: S = a(1 − r^n)/(1 − r). The division is legitimate only when 1 − r ≠ 0, that is when r ≠ 1. That is the condition in the standard, and it is not a technicality. If r = 1, every term equals a, the series is simply na, and the formula's denominator would be 0. The derivation itself shows why: with r = 1 the lines S and rS are identical and subtracting them gives 0 = 0, which tells us nothing. Check the formula on the small case: a = 3, r = 2, n = 8 gives 3(1 − 2⁸)/(1 − 2) = 3(1 − 256)/(−1) = 3 × 255 = 765, matching the direct sum. When r > 1 it is often tidier to write the equivalent form S = a(r^n − 1)/(r − 1), obtained by multiplying numerator and denominator by −1.

Using it for a savings total. Deposit 200 dollars at the end of each month into an account paying 0.25% per month, for 24 months. The first deposit earns interest for 23 months and grows to 200(1.0025)^23; the last deposit, made at the end, earns nothing and is worth 200. The total is 200 + 200(1.0025) + … + 200(1.0025)^23, a geometric series with a = 200, r = 1.0025, n = 24. So the total is 200(1.0025^24 − 1)/(1.0025 − 1) = 200(1.061757 − 1)/0.0025 ≈ 200 × 24.7028 ≈ 4,940.56 dollars. The deposits themselves total 4,800, so about 140.56 dollars is interest. Reporting to the cent is appropriate because the inputs are exact dollar amounts and an exact rate.

A worked model inside the lesson: the loan payment. A loan of L dollars is repaid by n equal monthly payments of M dollars, with interest charged at a monthly rate i on the balance. What must M be? Think about what each payment is worth today. A payment of M made one month from now is worth M/(1 + i) today, because that amount would grow to M in one month; a payment k months from now is worth M/(1 + i)^k today. The loan is fair when the present values of all n payments add up to L:

L = M/(1 + i) + M/(1 + i)² + … + M/(1 + i)^n.

This is a geometric series with first term a = M/(1 + i), ratio r = 1/(1 + i), and n terms. For a positive interest rate i > 0 the ratio is less than 1, so it is not 1 and the formula applies (at i = 0 the ratio is 1, and the payment is simply M = L/n): L = a(1 − r^n)/(1 − r) = [M/(1 + i)] · [1 − (1 + i)^(−n)] / [1 − 1/(1 + i)]. The denominator simplifies: 1 − 1/(1 + i) = i/(1 + i). Dividing by it multiplies by (1 + i)/i, and the (1 + i) cancels with the one in a: L = M[1 − (1 + i)^(−n)]/i. Solve for the payment: M = L·i / [1 − (1 + i)^(−n)].

Now use it on a hypothetical car loan: L = 10,000 dollars, nominal annual rate 6% compounded monthly, so i = 0.06/12 = 0.005 per month, n = 36 payments. (1.005)^36 ≈ 1.196681, so (1.005)^(−36) ≈ 0.835645, and 1 − 0.835645 = 0.164355. Then M = 10,000 × 0.005 / 0.164355 = 50 / 0.164355 ≈ 304.22 dollars per month. Total paid: 36 × 304.22 = 10,951.92 dollars, so the interest cost is about 951.92 dollars. Check the logic another way: if the balance is multiplied by 1.005 and reduced by 304.22 each month, starting from 10,000, it ends at about −0.02 dollars after the 36th payment: a 2-cent overpayment, because the exact payment 304.2194 was rounded up to 304.22.

A mortgage-style payment. The same formula with L = 200,000, i = 0.005 and n = 360 (30 years of monthly payments at a hypothetical 6% nominal annual rate, compounded monthly) gives M = 1,000/[1 − 1.005^(−360)] ≈ 1,000/0.833958 ≈ 1,199.10 dollars per month. Over 360 months that totals about 431,676 dollars, so the interest over the life of this hypothetical loan is about 231,676 dollars, more than the amount borrowed. Reading the formula explains why: with n = 360, the term (1 + i)^(−n) is small, so the denominator is close to 1 and the payment, 1,199.10 dollars, is only about 200 dollars more than the first month's interest on the full balance, L·i = 1,000 dollars. For many years most of each payment goes to interest, which is why the balance falls slowly at first. The rates here are chosen for arithmetic, not quoted from any lender.

What the formula cannot do and common errors. It applies only when the terms are genuinely geometric: a fixed ratio, every term. A series whose ratio changes, or an arithmetic series such as 3 + 5 + 7, needs a different tool. The exponent is n, the number of terms, not n − 1; students who add 1 + 2 + 4 + 8 (four terms, n = 4) sometimes use 2³ and get 7 instead of 15. And in the payment formula the rate must be the rate per payment period, so a nominal annual rate compounded monthly (the usual way loan rates are quoted) must be divided by 12 for monthly payments before anything else is done; an effective annual rate r would instead give the monthly rate (1 + r)^(1/12) − 1.

About the numbers. Every price, fee, amount and business in this lesson's examples, tasks and questions is made up for practice; none describes a real company, product or offer.

Worked derivation check on a small series: Find 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 with the formula and by direct addition. Here a = 3, r = 2, n = 8. S = a(r^n − 1)/(r − 1) = 3(2⁸ − 1)/(2 − 1) = 3 × 255 = 765. Direct addition: 3 + 6 = 9, + 12 = 21, + 24 = 45, + 48 = 93, + 96 = 189, + 192 = 381, + 384 = 765. The two agree, and the derivation's subtraction can be seen concretely: 2S = 6 + 12 + … + 768, and 2S − S = 768 − 3 = 765.

Worked growth-factor series: Find 100 + 100(1.05) + 100(1.05)² + … + 100(1.05)⁹, ten terms with a = 100 and r = 1.05. S = 100(1.05^10 − 1)/(1.05 − 1) = 100(1.628895 − 1)/0.05 = 100 × 12.5779 ≈ 1,257.79. Ten terms of 100 without growth would total 1,000, so the growth factor adds about 257.79. The exponent is 10 because there are ten terms, even though the largest exponent inside the series is 9.

Worked loan payment: A hypothetical loan of 10,000 dollars at 6% annual interest, compounded monthly, is repaid in 36 equal monthly payments. Monthly rate i = 0.06/12 = 0.005. M = L·i/[1 − (1 + i)^(−n)] = 10,000 × 0.005/[1 − 1.005^(−36)]. Compute 1.005^36 ≈ 1.196681, so 1.005^(−36) ≈ 0.835645 and the denominator is 0.164355. M ≈ 50/0.164355 ≈ 304.22 dollars per month. Total paid 36 × 304.22 ≈ 10,951.92 dollars; interest ≈ 951.92 dollars. Running the balance month by month, multiplying by 1.005 and subtracting 304.22, ends at about −0.02 dollars after 36 payments, zero except for the 2-cent effect of rounding the payment to the cent.

Worked series of fractions: Find 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 exactly. a = 1, r = 1/2, n = 7. S = (1 − (1/2)⁷)/(1 − 1/2) = (1 − 1/128)/(1/2) = 2(127/128) = 127/64 = 1.984375. Direct check: 64/64 + 32/64 + 16/64 + 8/64 + 4/64 + 2/64 + 1/64 = 127/64. The sum is just short of 2, and the formula shows why: S = 2 − 2(1/2)⁷ = 2 − 1/64.

Derivation record: Write out the derivation of S = a(1 − r^n)/(1 − r) in your own hand for the specific series 5 + 5(3) + 5(3)² + 5(3)³ + 5(3)⁴, showing the two lines S and 3S with terms aligned, the subtraction, the factoring and the division. State the value (5 × 242/2 = 605) and confirm it by direct addition: 5 + 15 + 45 + 135 + 405 = 605. Then write one sentence explaining what goes wrong in the derivation when r = 1, and compute 5 + 5 + 5 + 5 + 5 without the formula.

Independent performance: Design a hypothetical loan of your own: choose a principal, a nominal annual rate compounded monthly, and a number of monthly payments (for example 8,000 dollars at 4.8% for 24 months). Derive the payment formula from the present-value series, stating the first term, the ratio and the number of terms, then compute the payment, the total paid and the total interest. Build a month-by-month balance table for at least the first six months and the last two months (balance × (1 + i) − payment), and confirm the balance ends at zero within rounding. Finally explain, in a paragraph, why doubling the number of payments does not halve the payment. Success criteria: the monthly rate is the nominal annual rate divided by 12 before any other step, the exponent in the formula equals the number of payments, the balance table closes at zero when the unrounded payment is carried through it (with a payment rounded to the cent, a small leftover remains, a little over $1 for a 30-year, $200,000 loan at a nominal 6% paid at $1,199.10, and it is settled by adjusting the final payment, which the table shows), and the paragraph uses the size of (1 + i)^(−n) in its reasoning.