Multiplying out (a + b)ⁿ means expanding 2ⁿ products and collecting them. The binomial theorem writes the answer down directly: the coefficients are the binomial coefficients C(n, k), and they are there because expanding brackets is a counting problem.
Core ideas
The formula is short and the reason behind it is shorter: every coefficient is a count of something.
(a + b)ⁿ is n brackets multiplied together. Expanding means choosing a or b from each bracket in every possible way; a product with exactly k factors of b comes from choosing which k brackets supplied it, and there are C(n, k) ways to do that. Pascal's triangle appears here because it is a table of those counts.
The term is aⁿ⁻ᵏbᵏ, and (n - k) + k = n. It is a free check on every line you write: a³b³ cannot appear in the expansion of (a + b)⁵. There are n + 1 terms, one for each k from 0 to n.
(a - b)ⁿ is (a + (-b))ⁿ, so the term is C(n,k)aⁿ⁻ᵏ(-b)ᵏ and the sign is (-1)ᵏ. The signs alternate starting from +, and the terms with an even power of b keep their sign.
The material
The coefficients, the formula that produces them, and the reason they are counts.
Polynomials and Factoring expanded (a + b)² and (a + b)³ by hand. (a + b)⁵ is possible with patience, and (a + b)¹² is not: multiplying out twelve brackets means 2¹² = 4096 products before any collecting starts. The theorem below skips all of it.
Stripping out the coefficients leaves a triangle in which every entry is the sum of the two directly above it, with 1s down both edges. It is called Pascal's triangle, and row n holds the coefficients of (a + b)ⁿ.
For integers 0 ≤ k ≤ n, C(n, k) = n!/(k!(n - k)!), read "n choose k". It is also written with n above k in a single pair of brackets. The factorials are the ones from “Sequences and Recursion”, with 0! = 1.
C(9,4) = 9!/(4!5!). The 5! cancels most of the 9!, leaving (9·8·7·6)/(4·3·2·1) = 3024/24 = 126. Four factors on top because k = 4, and four below. The cancelling is not a shortcut for the lazy: 9! is 362880, and there is no reason to write it down.
For any numbers a, b and any nonnegative integer n, (a + b)ⁿ = Σ C(n,k) aⁿ⁻ᵏ bᵏ, the sum running over k from 0 to n. Written out, that is aⁿ + C(n,1)aⁿ⁻¹b + C(n,2)aⁿ⁻²b² + … + C(n,n-1)abⁿ⁻¹ + bⁿ.
Why the coefficient of aⁿ⁻ᵏbᵏ is C(n,k): expanding n brackets means taking one letter from each and multiplying, then adding every such product. A product ends up as aⁿ⁻ᵏbᵏ exactly when the b was taken from k of the brackets, and the number of ways to choose which k is C(n,k). Every one of those choices contributes 1, so they add to C(n,k).
That is also why the triangle's rule works: a choice of k brackets out of n either uses the last bracket, leaving k - 1 to choose from the first n - 1, or does not, leaving k to choose from n - 1. So C(n,k) = C(n-1,k-1) + C(n-1,k), which is the sum-of-the-two-above rule stated in symbols. The triangle is also symmetric, because C(n,k) = C(n,n-k): choosing k brackets to give b is the same act as choosing the other n - k to give a.
(a - b)⁴ = a⁴ - 4a³b + 6a²b² - 4ab³ + b⁴. The coefficients are row 4 unchanged; the signs come from (-b)ᵏ, so terms with odd k are negative. The last term is (-b)⁴ = +b⁴, which is the one people expect to be negative.
Setting a = b = 1 turns the theorem into 2ⁿ = C(n,0) + C(n,1) + … + C(n,n): each row of Pascal's triangle adds up to a power of two. Row 5 is 1, 5, 10, 10, 5, 1, totalling 32. It is a check on a row you have just written, and it is also the statement that a set of n things has 2ⁿ subsets.
Interactive
Pascal's triangle against the C(n, k) formula, against multiplying it out
Method
Four steps. The third is the one that decides whether the signs come out right.
In (2x - 3)⁴: a = 2x, b = -3, n = 4. The minus sign belongs to b, not to the expansion. Deciding this on paper before expanding is what keeps the signs honest.
Row n of Pascal's triangle if n is small, C(n,k) otherwise. For n = 4 they are 1, 4, 6, 4, 1, and the row is symmetric, which halves the work.
aⁿ⁻ᵏbᵏ for k = 0, 1, …, n, exponents totalling n in every term. Getting the skeleton down first stops a dropped term in the middle of a long expansion.
(2x)³ is 8x³, not 2x³, and (-3)² is 9, not -9. Then collect. Finally substitute x = 1: the expansion should equal the original expression evaluated there.
Both parts are compound: a carries a coefficient and b carries a sign. Every error in this expansion is one of those two being applied to only part of the term.
The check costs one line and catches almost everything, because substituting x = 1 turns the expansion into the sum of its coefficients. A dropped sign, a missing term or a mishandled (2x)³ will all show up in it.
Notice that the coefficients 16, -96, 216, -216, 81 are not symmetric even though row 4 is. The symmetry belongs to C(n,k); (2x) and (-3) are different sizes, and the powers of each pull the numbers out of balance.
For a faded expansion, take (x - 2)⁵. The supplied setup is a = x, b = -2, n = 5, with row 5 coefficients 1, 5, 10, 10, 5, 1. Write all six unsimplified terms before collecting their signs and powers, then substitute x = 1 into both the original bracket and your complete expansion before opening the quiz.
You should be able to produce row 6 of Pascal's triangle from row 5, compute a binomial coefficient by cancelling factorials, and expand (2x - 3)⁴ correctly including signs. For a complete expansion, the x = 1 check should be automatic.
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