The Binomial Theorem

Expanding (a + b)⁵ by hand is miserable. The binomial theorem gives it directly.

The coefficients are row n of Pascal's triangle, where each entry is the sum of the two above it:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

For (a + b)⁴: 1a⁴ + 4a³b + 6a²b² + 4ab³ + 1b⁴.

The pattern in the letters: a's exponent counts down from n, b's counts up from 0, and every term's exponents add to n.

With a subtraction, the signs alternate, because (a − b)ⁿ is (a + (−b))ⁿ.

The theorem states

(a + b)ⁿ = Σ_{k=0}^{n} C(n,k) a^{n−k} b^k

where C(n,k) = n!/(k!(n−k)!) is the binomial coefficient, the number of ways to choose k factors of b from n brackets. That combinatorial reading explains why the coefficients are exactly the combination counts and why Pascal's identity C(n,k) = C(n−1,k−1) + C(n−1,k) produces the triangle.

The expansion has n + 1 terms, and the (k+1)th term is C(n,k)a^{n−k}b^k — the standard way to extract a single requested term without expanding everything.

Substituting into the theorem yields useful identities: a = b = 1 gives Σ C(n,k) = 2ⁿ, the number of subsets of an n-element set.

For a binomial like (2x − 3)⁴, the whole of 2x and −3 must be substituted for a and b, so coefficients pick up 2^{n−k} and (−3)^k factors — the most common source of error in applying the theorem.