DWG NO. 9.5 — Lesson 5 of 6

The Binomial Theorem

Unit 9: Sequences, Series, and Combinatorics · ~25 min

Objective Expand a binomial power (a + b\()^{n}\) using the Binomial Theorem and Pascal's Triangle, and find a single specified term without expanding the whole thing.

Multiplying out (a + b\()^{2}\) = a² + 2ab + b² by hand is routine. (a + b\()^{7}\) is not — but its expansion follows a completely predictable pattern in both its terms and their coefficients. The Binomial Theorem gives that pattern directly, using the counting tools this unit is building toward in Lesson 9.6.

Binomial coefficients

The coefficient of the term \(a^{n-k} b^{k}\) in the expansion of (a+b\()^{n}\) is written _n\(C_{k}\) (read "n choose k") and defined as:

Binomial coefficient _n\(C_{k}\) = \(\frac{n!}{k!(n-k)!}\)

where n! ("n factorial") means n(n−1)(n−2)…(2)(1), with 0! defined as 1. These same numbers form Pascal's Triangle: each row starts and ends with 1, and every interior entry is the sum of the two entries above it. Row n of the triangle (starting the count at row 0) gives the coefficients for (a+b\()^{n}\).

Pascal's Triangle — row 4 (1, 4, 6, 4, 1) gives the coefficients of (a+b)⁴

The theorem

Binomial Theorem (a + b\()^{n}\) = \(\sum_{k=0}^{n}\) _n\(C_{k}\) \(a^{n-k} b^{k}\)

Reading the pattern: the exponent on a starts at n and counts down to 0 while the exponent on b starts at 0 and counts up to n, and every term's two exponents always add to n. Each term's coefficient is the corresponding entry from Pascal's Triangle.

Finding a single term

Sometimes only one term is needed, not the full expansion. The (k+1)th term of (a+b\()^{n}\) (counting the first term as k=0) is:

General term \(T_{k+1}\) = _n\(C_{k}\) \(a^{n-k} b^{k}\)

Worked Example 1 · Full expansion

ProblemExpand (x + 2\()^{4}\) using the Binomial Theorem.

1Coefficients from row 4 of Pascal's Triangle: 1, 4, 6, 4, 1.

2(x+2\()^{4}\) = 1·\(x^{4}\)(2\()^{0}\) + 4·\(x^{3}\)(2\()^{1}\) + 6·\(x^{2}\)(2\()^{2}\) + 4·\(x^{1}\)(2\()^{3}\) + 1·\(x^{0}\)(2\()^{4}\).

3= \(x^{4}\) + 8\(x^{3}\) + 24\(x^{2}\) + 32x + 16.

x⁴ + 8x³ + 24x² + 32x + 16

Worked Example 2 · A single term

ProblemFind the term containing x³ in the expansion of (2x − 1\()^{5}\).

1Here a = 2x, b = −1, n = 5. The exponent on a is 5−k = 3, so k = 2.

2\(T_{3}\) = ₅\(C_{2}\) (2x\()^{3}\)(−1\()^{2}\) = 10 · 8x³ · 1.

80x³

Guided practice

1.Write row 5 of Pascal's Triangle.
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1, 5, 10, 10, 5, 1.
2.Expand (x + 1\()^{3}\) using the Binomial Theorem.
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Row 3: 1,3,3,1. x³ + 3x² + 3x + 1.
3.Expand (x − 3\()^{4}\).
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Row 4: 1,4,6,4,1, with alternating signs from (−3): x⁴ − 12x³ + 54x² − 108x + 81.
4.Find the coefficient of a²b⁵ in the expansion of (a + b\()^{7}\).
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Need exponent on a to be 2, so k=5. ₇\(C_{5}\) = 7!/(5!2!) = 21.
5.Find the term containing x² in the expansion of (x + 3\()^{6}\).
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Exponent on x is 6−k = 2, so k=4. \(T_{5}\) = ₆\(C_{4}\) x²(3)⁴ = 15 · x² · 81 = 1215x².

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