Binomial Theorem

The binomial theorem expands a power of a binomial without multiplying it out by hand: (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. The coefficients \binom{n}{k} come from combinations or from row n of Pascal’s triangle. The powers of a go down from n to 0 while the powers of b go up from 0 to n.

Before this lesson: Permutations and Combinations, Special Products of Binomials and Sigma Notation

The pattern behind the binomial theorem

Multiply out a few powers of (a + b) and a pattern appears:

(a + b)^2 = a^2 + 2ab + b^2

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

Each power has n + 1 terms. The exponent on a starts at n and drops by 1 each term, and the exponent on b starts at 0 and rises by 1. In every term the two exponents add to n.

The coefficients 1, 2, 1 and 1, 3, 3, 1 are rows of Pascal’s triangle, where each number is the sum of the two numbers above it. They are also the combinations \binom{n}{k}, which count how many ways to pick b from k of the n factors.

The binomial theorem formula

(a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n

The binomial coefficient is \binom{n}{k} = \frac{n!}{k!\,(n - k)!}, the same as C(n, k).

To find one term without the whole expansion, use the general term. The term with b^k is term number k + 1:

\text{term } k + 1 = \binom{n}{k} a^{n-k} b^k

Binomial theorem examples

Example 1: Expand with Pascal’s triangle

Expand (x + 2)^4.

(x + 2)^4 = 1x^4 + 4x^3(2) + 6x^2(2)^2 + 4x(2)^3 + 1(2)^4

(x + 2)^4 = x^4 + 4x^3(2) + 6x^2(4) + 4x(8) + 16

(x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16

Answer: x^4 + 8x^3 + 24x^2 + 32x + 16.

Example 2: A binomial with a coefficient and a minus sign

Expand (2x - 3)^3.

(2x - 3)^3 = 1(2x)^3 + 3(2x)^2(-3) + 3(2x)(-3)^2 + 1(-3)^3

(2x - 3)^3 = 8x^3 + 3(4x^2)(-3) + 3(2x)(9) + (-27)

(2x - 3)^3 = 8x^3 - 36x^2 + 54x - 27

Answer: 8x^3 - 36x^2 + 54x - 27.

Example 3: Find one term

Find the 4th term of (x + 3)^7.

k = 3

\binom{7}{3} x^{7-3} (3)^3

\binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35

35 \cdot x^4 \cdot 27

945x^4

Answer: 945x^4.

Example 4: Find a coefficient

Find the coefficient of x^5 in (2x - 1)^8.

\binom{8}{k}(2x)^{8-k}(-1)^k

8 - k = 5

k = 3

\binom{8}{3}(2x)^5(-1)^3

56 \cdot 32x^5 \cdot (-1)

-1792x^5

Answer: the coefficient is −1792.

Example 5: Word problem with savings growth

A savings account grows 2% per year, so $1,000 becomes 1000(1.02)^{10} dollars after 10 years. Use the first three terms of the binomial theorem to estimate it.

(1 + 0.02)^{10} \approx \binom{10}{0} + \binom{10}{1}(0.02) + \binom{10}{2}(0.02)^2

(1 + 0.02)^{10} \approx 1 + 10(0.02) + 45(0.0004)

(1 + 0.02)^{10} \approx 1 + 0.2 + 0.018

(1 + 0.02)^{10} \approx 1.218

1000(1.218) = 1218

Answer: about $1,218. The exact value is $1,218.99, so three terms come within about a dollar, because the later terms use tiny powers of 0.02. For more on growth like this, see compound interest.

Common mistakes

Related methods

Building Pascal’s triangle

Start with a 1 on top. Each row begins and ends with 1, and every other number is the sum of the two numbers just above it. This works because \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}.

Sum of the coefficients

Setting every variable equal to 1 adds up all the coefficients. For (a + b)^n that sum is 2^n, which is also the sum of row n of Pascal’s triangle.

Practice problems

1. Write row 5 of Pascal’s triangle.
2. Expand (x + 1)^5.
3. Expand (a - b)^4.
4. Expand (3x + 2)^3.
5. Expand (x^2 + 1)^3.
6. Find the 3rd term of (x + 5)^6.
7. Find the coefficient of x^3 in (x - 2)^7.
8. Find the constant term of \left(x + \frac{1}{x}\right)^6.
9. Find the sum of the coefficients of (2x + 1)^5.
10. Use the first three terms of the binomial theorem to estimate 0.99^6.

1, 4, 6, 4, 1

1, 5, 10, 10, 5, 1

(x + 1)^5 = x^5 + 5x^4(1) + 10x^3(1)^2 + 10x^2(1)^3 + 5x(1)^4 + (1)^5

(x + 1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1

(a - b)^4 = a^4 + 4a^3(-b) + 6a^2(-b)^2 + 4a(-b)^3 + (-b)^4

(a - b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4

(3x + 2)^3 = (3x)^3 + 3(3x)^2(2) + 3(3x)(2)^2 + (2)^3

(3x + 2)^3 = 27x^3 + 3(9x^2)(2) + 3(3x)(4) + 8

(3x + 2)^3 = 27x^3 + 54x^2 + 36x + 8

(x^2 + 1)^3 = (x^2)^3 + 3(x^2)^2(1) + 3(x^2)(1)^2 + (1)^3

(x^2 + 1)^3 = x^6 + 3x^4 + 3x^2 + 1

\binom{6}{2}x^{4}(5)^2

15 \cdot x^4 \cdot 25

375x^4

7 - k = 3

k = 4

\binom{7}{4}(-2)^4 = 35 \cdot 16

\binom{7}{4}(-2)^4 = 560

\binom{6}{k}x^{6-k}\left(\frac{1}{x}\right)^k = \binom{6}{k}x^{6-2k}

6 - 2k = 0

k = 3

\binom{6}{3} = 20

(2(1) + 1)^5 = 3^5

3^5 = 243

(1 - 0.01)^6 \approx 1 + 6(-0.01) + 15(-0.01)^2

(1 - 0.01)^6 \approx 1 - 0.06 + 0.0015

(1 - 0.01)^6 \approx 0.9415

Binomial theorem FAQ

What is the binomial theorem used for?

It expands powers like (a + b)^n quickly and finds a single term or coefficient without the full expansion. It also helps estimate powers such as 1.02^{10}.

How is Pascal’s triangle related to the binomial theorem?

Row n of Pascal’s triangle lists the coefficients of (a + b)^n. Each entry equals \binom{n}{k}.

How many terms are in the expansion of (a + b)^n?

There are n + 1 terms. For example, (a + b)^4 has 5 terms.

How do you find a specific term in a binomial expansion?

The term with b^k is \binom{n}{k}a^{n-k}b^k, and it is term number k + 1. Set the exponent you want equal to the right power and solve for k.

Summary

• (a + b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k, with n + 1 terms.
• Coefficients come from row n of Pascal’s triangle or from \binom{n}{k}.
• Powers of a count down from n; powers of b count up from 0; each term’s exponents add to n.
• Keep parentheses around a and b, and use a negative b for a minus sign.
• Term k + 1 is \binom{n}{k}a^{n-k}b^k.

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