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.
Comparing the coefficient of xn+r on both sides of (1+x)n(x+1)n=(1+x)2n, with the second factor written in decreasing powers of x, connects the given sum to a single binomial coefficient. …
Multiply (1+x)n (written in increasing powers) by (x+1)n (written in decreasing powers) and compare the coefficient of xn+r on both sides with (1+x)2n.
Write (1+x)n=C0+C1x+C2x2+⋯+Cnxn ... (i)
Also (x+1)n can be written with powers of x in decreasing order: the coefficient of xn−k is Ck, i.e. (x+1)n=C0xn+C1xn−1+⋯+Cn ... (ii)
Since (1+x)n(x+1)n=(1+x)2n, multiply (i) and (ii) and compare the coefficient of xn+r on both sides.
On the right, the coefficient of xn+r in (1+x)2n is 2nCn+r.
On the left, using (i) and (ii): the coefficient of xm in (ii) is Cn−m. A term xk from (i) (coefficient Ck) pairs with xn+r−k from (ii) (coefficient Cn−(n+r−k)=Ck−r), valid for k≥r. So the coefficient of xn+r on the left is: