Lesson 10: Series and the Binomial Theorem

A series is a sum of a sequence, and summing infinitely many terms — sensibly — is the second great limit idea of precalculus. This lesson covers arithmetic and geometric series (finite and infinite) and the binomial theorem, which expands (a + b)^n exactly. Both topics are algebra on the surface and limits underneath: the infinite geometric series only exists because a limit does.

Arithmetic series

An arithmetic sequence has a common difference d: a, a + d, a + 2d, ... The nth term is a_n = a + (n − 1)d, and the sum of n terms pairs first with last: S_n = (n/2)(a + a_n) = (n/2)(2a + (n − 1)d). Worked example: 3 + 7 + 11 + ... for 20 terms. Here a = 3, d = 4, a_20 = 3 + 19·4 = 79, so S_20 = (20/2)(3 + 79) = 10 × 82 = 820. The pairing argument — write the sum forwards and backwards and add — is Gauss's schoolroom trick, and the formula is just that trick recorded. Watch the term count: from 3 to 79 in steps of 4 there are (79 − 3)/4 + 1 = 20 terms; the "+1" for the first term is the most-lost mark in the topic.

Geometric series, finite

A geometric sequence has a common ratio r: a, ar, ar^2, ... with a_n = a·r^(n − 1) and S_n = a(1 − r^n)/(1 − r). Worked example: 5 + 5/2 + 5/4 + ... for 8 terms: a = 5, r = 1/2, so S_8 = 5(1 − (1/2)^8)/(1 − 1/2) = 10(1 − 1/256) = 10 × 255/256 ≈ 9.961. Note the sum is already close to something — the terms are dying fast. The derivation matters as much as the formula: multiply S_n by r, subtract, and everything telescopes except the first and last terms. That telescoping idea returns in integration and in probability.

Infinite geometric series: a limit in disguise

If |r| < 1, then r^n → 0 as n → ∞, and the finite formula leaves S = a/(1 − r). Worked example: 9 + 3 + 1 + 1/3 + ... = 9/(1 − 1/3) = 9/(2/3) = 13.5. Sanity check with partial sums: 9, 12, 13, 13.33, 13.44 — closing in on 13.5 from below ✓. If |r| ≥ 1 the series diverges: the partial sums never settle. The repeating decimal 0.777... is the series 7/10 + 7/100 + ... = (7/10)/(1 − 1/10) = 7/9 — every repeating decimal is a fraction because every repeating decimal is a convergent geometric series. Zeno's paradox dissolves the same way: infinitely many steps, finite total, because the ratio is less than 1.

The binomial theorem

(a + b)^n expands into n + 1 terms whose coefficients are the binomial coefficients C(n, k) = n!/(k!(n − k)!) — row n of Pascal's triangle. The general term is C(n, k)·a^(n − k)·b^k. Worked example: expand (2x − 3)^4.

k=0: 1 · (2x)^4        = 16x^4
k=1: 4 · (2x)^3 · (−3) = −96x^3
k=2: 6 · (2x)^2 · 9    = 216x^2
k=3: 4 · (2x) · (−27)  = −216x
k=4: 1 · 81            = 81

So (2x − 3)^4 = 16x^4 − 96x^3 + 216x^2 − 216x + 81. Verify at x = 1: (2 − 3)^4 = 1, and 16 − 96 + 216 − 216 + 81 = 1 ✓ — always spot-check an expansion at a convenient value. Finding a single term: the coefficient of x^3 in (x + 2)^7 comes from the term C(7, k)·x^(7 − k)·2^k with 7 − k = 3, i.e. k = 4: C(7, 4)·2^4 = 35 × 16 = 560. Target the exponent first, then assemble the coefficient — never expand the whole power to hunt for one term.

Modelling: the bouncing ball

A ball dropped from 12 m rebounds to 2/3 of its previous height each bounce. The total distance travelled until it (theoretically) rests: the first drop is 12; then each bounce contributes an up *and* a down: 2 × 12 × (2/3) + 2 × 12 × (2/3)^2 + ... Factoring: 12 + 24·[(2/3)/(1 − 2/3)] = 12 + 24 × 2 = 60 m. Two subtleties carry the marks: the first drop is counted once (it is not part of the symmetric up-down series), and the infinite series converges because r = 2/3 < 1. The height after the 5th bounce is 12(2/3)^5 = 12 × 32/243 ≈ 1.58 m — geometric sequences describe the heights, geometric series the distances.

Common mistakes

• Miscounting terms. S_n formulas need the number of terms, not the last index you happened to notice. From the 3rd to the 20th term inclusive is 18 terms.
• Using r instead of r^n in the finite formula. S_8 above needs (1/2)^8, the ratio raised to the *number of terms* — the formula is a(1 − r^n)/(1 − r), and the exponent is n, not n − 1, because a_n's exponent (n − 1) is a different count.
• Applying a/(1 − r) when |r| ≥ 1. 9/(1 − 3) = −4.5 is not the sum of 9 + 27 + 81 + ...; divergence means no finite sum exists. Check |r| first.
• Losing the sign in binomial terms. In (2x − 3)^4 the "b" is −3, so odd-k terms are negative. Substituting b = 3 and sprinkling signs afterwards invites errors.
• Exponent drift in the general term. In C(n, k)·a^(n − k)·b^k the exponents sum to n in every term. If yours do not (e.g. x^3 with 2^4 in a power-7 expansion: 3 + 4 = 7 ✓), the term is wrong.

Exam Experience

1. The sum of the first 20 terms of the arithmetic series 5 + 9 + 13 + ... is

• (A) 810
• (B) 840
• (C) 860
• (D) 900

Answer: (C). d = 4, a_20 = 5 + 19·4 = 81, S_20 = (20/2)(5 + 81) = 10 × 86 = 860. (A) uses a_20 = 77 — the (n − 1) forgotten — and (D) pairs 5 with 85 (one term too many).

2. The infinite series 6 + 2 + 2/3 + 2/9 + ... converges to

• (A) 8
• (B) 9
• (C) 10
• (D) it diverges

Answer: (B). a = 6, r = 1/3, |r| < 1, so S = 6/(1 − 1/3) = 6/(2/3) = 9. Partial sums 6, 8, 8.67, 8.89 approach 9 from below ✓.

3. The coefficient of x^2 in the expansion of (3x − 1)^5 is

• (A) 90
• (B) −90
• (C) 270
• (D) −270

Answer: (B). The x^2 term has k = 3 (so that (3x)^(5 − 3) = (3x)^2): C(5, 3)·(3x)^2·(−1)^3 = 10 × 9x^2 × (−1) = −90x^2. (A) drops the sign from (−1)^3; (D) misapplies the coefficient as C(5, 3)·27.

Constructed response. A ball is dropped from a height of 12 m. Each bounce reaches 2/3 of the height of the previous fall. (a) How high does the ball rise after the 1st bounce, and after the 5th bounce? (b) Find the total distance travelled by the ball from release until it comes to rest, treating the bounces as continuing indefinitely. (c) A second ball is dropped from 12 m but rebounds to 3/2 of... explain, in one or two sentences, why the method of part (b) cannot be applied if the rebound ratio were greater than 1.

Scoring guide. (a) 2 points: after the 1st bounce, 12 × 2/3 = 8 m — 1 point; after the 5th, 12(2/3)^5 = 384/243 ≈ 1.58 m — 1 point. (b) 3 points: up-and-down series after the first drop has a = 2 × 12 × (2/3) = 16 and r = 2/3 — 1 point; its sum is 16/(1 − 2/3) = 48 — 1 point; total = 12 + 48 = 60 m — 1 point. (c) 1 point: with ratio ≥ 1 the terms do not shrink to zero, so the infinite series diverges and no finite total distance exists (the ball would gain energy, which is also physically impossible).

Going deeper: saving as a geometric series, and telescoping

Regular deposits. Deposit $100 at the end of each month into an account paying 0.5% monthly. After 12 months the first deposit has grown for 11 months, the last for 0: FV = 100(1.005^11 + 1.005^10 + ... + 1) — a geometric series with a = 100, r = 1.005, n = 12 terms, read backwards. FV = 100 × (1.005^12 − 1)/(1.005 − 1) = 100 × (0.061678)/(0.005) ≈ $1,233.56: you deposited $1,200 and the series formula priced the interest at $33.56. Every annuity, mortgage and pension calculation in finance is this formula with different letters — geometric series are not an exam topic, they are plumbing.

Telescoping series collapse from inside. Since 1/(k(k + 1)) = 1/k − 1/(k + 1), the sum Σ_{k=1}^{n} 1/(k(k + 1)) = (1 − 1/2) + (1/2 − 1/3) + ... + (1/n − 1/(n + 1)) = 1 − 1/(n + 1): every interior term cancels. For n = 9 the sum is 0.9 exactly, and the infinite series converges to 1. The partial-fractions move from Lesson 15 is what exposed the telescope — notice how topics in this course keep paying each other back. When a series resists the standard formulas, ask whether its terms secretly split into a difference; telescoping is summation's version of the factor-and-cancel trick from limits.

Workshop: finding n, finding terms

A later term and a long sum. For 4, 12, 36, ... (r = 3): a_10 = 4·3^9 = 4 × 19,683 = 78,732, and S_10 = 4(3^10 − 1)/(3 − 1) = 2 × 59,048 = 118,096. Inverting a series problem. For the arithmetic series 3 + 7 + 11 + ... with sum 820, find n: S_n = (n/2)(2·3 + (n − 1)·4) = n(2n + 1) = 820, so 2n^2 + n − 820 = 0 and n = (−1 + sqrt(1 + 6560))/4 = (−1 + 81)/4 = 20 (the negative root is rejected — n counts terms). Series problems run in both directions: given n find the sum, given the sum find n — the second direction is a quadratic in disguise, and Lesson 10's formulas are what make it visible.