--- title: "The Binomial Expansion" description: "For a nonnegative integer n, expand (a + b)^n using Pascal's triangle or nCr, find a single term or coefficient, and approximate (1 + h)^n using early terms when |h| is small, checking the omitted ter" canonical: https://lightmysky.com/learn/mathematics/the-binomial-expansion-mt_z16HZkNZ8c source: https://lightmysky.com/learn/mathematics/the-binomial-expansion-mt_z16HZkNZ8c.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # The Binomial Expansion For a nonnegative integer n, expand (a + b)^n using Pascal's triangle or nCr, find a single term or coefficient, and approximate (1 + h)^n using early terms when |h| is small, checking the omitted terms. Subject: Mathematics · Area: Algebra · Ages 16 to 17 Page: https://lightmysky.com/learn/mathematics/the-binomial-expansion-mt_z16HZkNZ8c ## Ready when they can - Expand (1 + 2x)⁵ in full - Find a stated term, its position or a coefficient without expanding everything - Approximate 1.02⁸ from the first three terms and comment on the error ## Lesson: Expanding brackets without the pain A binomial has two terms. For example, (a + b)^2 = (a + b)(a + b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2. Each product takes one part from each bracket. For a nonnegative integer n, terms of (a + b)^n are (n choose k) × a^(n-k) × b^k for k = 0 to n. Powers of a decrease as powers of b increase, giving n + 1 symbolic terms. Substituting values or collecting like terms may make terms vanish or combine. The number n choose k counts selections of k bracket positions without order. Choosing 2 of 5 gives 5 × 4 ordered picks, but each pair occurs twice, so the count is 10. A calculator's nCr function takes n and k and gives this count. Choosing none or all gives 1. With b's powers increasing, k = 0 gives the first term, so the term for k has position k + 1. *(drawing: For n = 4, the exponents add to 4 in every power pair. The coefficients still need to be included.)* Pascal's triangle gives the selection counts. Number the top row 0; it contains 1. Each new row starts and ends with 1. Add adjacent entries in the previous row to obtain the interior entries. Row 4 is 1, 4, 6, 4, 1. It produces row 5 below. Row n gives n choose k as k runs from 0 to n. **Example.** To expand (1 + cx)^n, multiply each entry in row n by the matching power (cx)^k. Raise both c and x to k. For (1 + 3x)^4, row 4 gives 1, 4, 6, 4, 1. The expansion is 1 + 12x + 54x^2 + 108x^3 + 81x^4. For example, its x^2 term is 6 × (3x)^2 = 54x^2. For (a + cx)^n, the coefficient of x^k is (n choose k) × a^(n-k) × c^k. Use nCr with n and k for the selection count, and include both constants. For (4 + x)^3, the x^2 term is 3 × 4 × x^2 = 12x^2. For (a - cx)^n, use -c instead of c. When c is positive, odd powers of -c are negative and even powers are positive. If the first part contains x, work out its exponent before choosing k. In (x^2 + 2)^4, term k contains x raised to 2(4-k). To locate x^2, solve 2(4-k) = 2, giving k = 3. This is term number 4 when terms are ordered by decreasing powers of x, because term k has position k + 1. For integer n at least 2, the first three terms of (1 + h)^n are 1 + nh + (n choose 2)h^2. A small |h| can make later terms small, but check them for the chosen n. For 1.03^5, use h = 0.03: 1 + 0.15 + 0.009 = 1.159. All omitted terms are positive here, so this is an underestimate. Adding them gives the exact value. For negative h, omitted terms can have different signs. Add them to decide whether an estimate is above or below the exact value. In (1 + h)^4, the first three terms leave out 4h^3 + h^4 = h^3(4 + h). If -1 < h < 0, this remainder is negative, so the three-term sum is an overestimate. **Recap.** Pair each coefficient with falling and rising powers, find single terms with n choose k, and approximate with the first terms. *This lesson, question bank and flashcards were written for this dataset.* ## Practice 14 questions on this page, each with its working shown. ## Needs first - [Counting Outcomes with the Product Rule](https://lightmysky.com/learn/mathematics/counting-outcomes-with-the-product-rule-mt_74gSbYIG8p) ## Optional connections - [Geometric Series and the Sum to Infinity](https://lightmysky.com/learn/mathematics/geometric-series-and-the-sum-to-infinity-mt_HHxFOv6vXS) ## Opens up - [Binomial Coefficients and Combinatorial Identities](https://lightmysky.com/learn/mathematics/binomial-coefficients-and-combinatorial-identities-mt_lHXFvudTh6) - [Taylor and Maclaurin Series](https://lightmysky.com/learn/mathematics/taylor-and-maclaurin-series-mt_SgI9Pn9RIO) - [The Binomial Distribution](https://lightmysky.com/learn/mathematics/the-binomial-distribution-mt_X2kVGK67Md)