Imagine you have to expand (x+y)2. You know it's x2+2xy+y2. What about (x+y)3? That's x3+3x2y+3xy2+y3. Now try (x+y)4 — you could multiply (x+y)3 by (x+y) again, but it gets messy fast.
The Binomial Theorem is the shortcut. It tells you exactly what (x+y)n expands to, for any positive integer n, without doing the multiplication step by step.
The Pattern You Already Know
Look at the expansions we have:
Power
Expansion
(x+y)0
1
(x+y)1
x+y
(x+y)2
x2+2xy+y2
(x+y)3
x3+3x2y+3xy2+y3
(x+y)4
x4+4x3y+6x2y2+4xy3+y4
Three things stand out:
The powers of x decrease from n down to 0, while the powers of y increase from 0 up to n. In each term, the exponents add to n.
The coefficients — 1, 2, 1 for n=2; 1, 3, 3, 1 for n=3; 1, 4, 6, 4, 1 for n=4 — follow a famous pattern called Pascal's triangle.
The number of terms is always n+1.
Note
Pascal's triangle: each number is the sum of the two numbers directly above it.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Why Do These Coefficients Appear?
Think about what (x+y)n really means. It's (x+y) multiplied by itself n times:
(x+y)n=n factors(x+y)(x+y)⋯(x+y)
When you expand, you pick either x or y from each factor. A term like xn−kyk comes from choosing y from exactly k of the n factors and x from the rest.
How many ways can you choose which k factors give you y? That's exactly the number of combinations: (kn) (read "n choose k").
(kn)=k!(n−k)!n!
So the coefficient of xn−kyk is (kn). That's the heart of the theorem.
Multiply (1+x)n by itself to get (1+x)2n, and compare the coefficient of xn computed two different ways — as a convolution on the left, and directly on the right.
Step 1. Set up the identity.(1+x)n⋅(1+x)n=(1+x)2n.
Step 2. Coefficient of xn on the LEFT. Using (1+x)n=∑k=0nCkxk for each factor, the coefficient of xn in the product is
∑k=0nCk⋅Cn−k
(pick xk from the first factor and xn−k from the second, for every k from 0 to n).
Step 3. Simplify using Cn−k=Ck. By the symmetry property nCn−k=nCk,