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The Binomial Theorem gives the expansion of (a + b)n: the sum from k=0 to n of C(n,k) * an-k * bk, where C(n,k) = n!/(k!(n-k)!) is the binomial coefficient. Pascal's Triangle provides these coefficients visually. The theorem generalizes FOIL to any positive integer power. Each term has degree n (the exponents of a and b sum to n), and there are n+1 terms total.
Start with manual expansion of (a+b)2, (a+b)3, (a+b)4 and observe patterns. Introduce Pascal's Triangle as the coefficient pattern. Formalize with the binomial coefficient formula. Practice expanding specific binomials and finding specific terms (e.g., "the 4th term of (2x - 3)7"). Connect to combinations.
You already know how to expand (a + b)² = a² + 2ab + b² and (a + b)³ = a³ + 3a²b + 3ab² + b³ by multiplying out. But what about (a + b)¹⁰ or (a + b)²⁰? Multiplying out twenty factors by hand is impractical, and the Binomial Theorem gives you a direct formula instead. The key insight is that the coefficients — 1, 2, 1 for (a+b)², and 1, 3, 3, 1 for (a+b)³ — are not arbitrary. They are binomial coefficients C(n,k), which you know from combinatorics as the number of ways to choose k items from n.
Here is why: when you expand (a + b)n by multiplying out n copies of (a + b), each term in the expansion comes from choosing either the a or the b from each factor. To get the term an-kbk, you must choose b from exactly k of the n factors and a from the remaining n-k. The number of ways to make that choice is C(n,k) = n!/(k!(n-k)!). So the term containing bk appears exactly C(n,k) times, and the coefficient of an-kbk in the expansion is C(n,k). The full expansion is the sum of all these terms as k runs from 0 to n.
The formula is (a + b)n = Σ C(n,k) · an-k · bk, summing from k=0 to n. Notice that the exponents of a and b always sum to n — every term has total degree n. There are n+1 terms (k = 0, 1, 2, ..., n). Pascal's Triangle arranges the coefficients visually: each row gives the coefficients for the corresponding power, and each entry is the sum of the two above it, which mirrors the identity C(n,k) = C(n-1,k-1) + C(n-1,k).
In practice, the Binomial Theorem is often used to find a specific term without expanding the whole polynomial. The (k+1)-th term of (a+b)n is C(n,k) · an-k · bk. For example, to find the 4th term (k=3) of (2x - 3)7, plug in a = 2x, b = -3, n = 7, k = 3: C(7,3) · (2x)4 · (-3)3 = 35 · 16x⁴ · (-27) = -15,120x⁴. The negative sign comes from b = -3 raised to an odd power — a common place to make errors. The theorem turns what was a tedious multiplication into a single lookup and calculation.