the binomial theorem

Everyone learns that (a + b)^2 = a^2 + 2ab + b^2. What about (a+b)^5, or (a+b)^20? Multiplying out by hand is hopeless, yet there is a clean pattern. The binomial theorem is the formula that tells you, for any whole number n, exactly what (a + b)^n expands to — and the coefficients turn out to be the binomial coefficients you already know from counting.

The statement is: (a + b)^n = sum from k=0 to n of C(n,k) a^(n-k) b^k. Spelled out, that is C(n,0) a^n + C(n,1) a^(n-1) b + C(n,2) a^(n-2) b^2 + ... + C(n,n) b^n. The exponents of a fall from n to 0 while the exponents of b rise from 0 to n, and each term's coefficient is 'n choose k'. The reason combinations appear is pure counting: (a+b)^n is n copies of (a+b) multiplied together, and to form the term with b^k you must pick the b from exactly k of the n factors and the a from the rest; the number of ways to choose which k factors give the b is C(n,k). For example (a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3, with coefficients 1, 3, 3, 1 — Pascal's row 3.

The theorem is a bridge between algebra and counting, and it is the source of many identities. Setting a = b = 1 gives sum of C(n,k) = 2^n (the subset count). Setting a = 1, b = -1 gives the alternating sum C(n,0) - C(n,1) + C(n,2) - ... = 0 for n >= 1. As stated here it is for non-negative integer n; Newton's generalised binomial series extends it to fractional and negative powers, but that version is an infinite series and needs convergence conditions, so do not assume the finite formula carries over unchanged.