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.
What a learner can do afterwards
- 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
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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.
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.
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.
Pair each coefficient with falling and rising powers, find single terms with n choose k, and approximate with the first terms.
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More to read
Counting bracket selections
Write (a + b)^n as n copies of the bracket, where n is a positive integer. Each product takes one part from every copy. Taking b from k copies leaves n - k copies supplying a. The resulting product is a^(n-k)b^k.
Different selections of bracket positions can give the same product. Their number is n choose k. To choose two positions out of seven, first count ordered picks: 7 × 6 = 42. Every pair appears in both orders, so divide by 2. There are 21 pairs.
A calculator's nCr function gives these selection counts. Pascal's triangle contains the same numbers. When powers of b increase, k starts at 0. Thus the term with index k is term number k + 1. The formula has n + 1 symbolic terms, though substituting values can make terms combine or vanish.
Keeping constants and signs
The entries in Pascal's triangle count selections. They do not include powers of the constants inside a substituted bracket. For (a + cx)^n, the coefficient of x^k is (n choose k) × a^(n-k) × c^k.
For example, expand (5 - 2x)^3 using row 3: 1, 3, 3, 1. The four terms are 5^3, 3 × 5^2 × (-2x), 3 × 5 × (-2x)^2 and (-2x)^3. Their sum is 125 - 150x + 60x^2 - 8x^3.
Keep -2 inside parentheses when raising it to a power. Its even powers are positive and its odd powers are negative. To find just one coefficient, calculate only that term, including the selection count and both constants' powers.
From multiplication to matching products
Start with (2 + y)^2 = (2 + y)(2 + y). Multiply each part of the first bracket by each part of the second. The products are 4, 2y, 2y and y^2. Collecting them gives 4 + 4y + y^2.
The two middle products match. One takes y from the first bracket; the other takes y from the second. There are two selections producing the same product. With more brackets, binomial coefficients count how often each product occurs.
Keep a fixed order when expanding. Increase the second part's exponent from 0 to n and decrease the first part's exponent from n to 0. Multiply each power pair by its selection count. Before substituting values or collecting terms, check that there are n + 1 terms and that each exponent pair adds to n.
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