Expert: Sherry Wallin - 8/8/2009

Question
Hi.
Can you give me a basic high school level explanation about what the Binomial theorem is and some examples?

Answer
A high school level explanation, hmmm...
Well one thing that can be done is any binomial to an integer power can be taken to the power without actually multiplying it out. For example (a+b)^2 = a^+2ab+b^2 which is elementary and so is (a+b)^3 = a^3+3a^2b+3ab^2+b^3, meaning you can multiply (a+b)(a+b)(a+b)and easily arrive at that answer. But what if you wanted to find (a+b)^5? This can become quite tedious but if you know Pascal's triangle or the binomial theorem you can figure out the coefficients for all the powers of a and b.
           1
        1    1
      1   2   1
    1   3   3   1
  1   4   6   4   1
1  5   10 10  5   1

This is known as Pascal's triangle and every new row is composed of the sum of the 2 numbers above it so for example in the 4th row the 6 comes from 3+3 and in the 5th row the 5 comes from 1+4 as well as the 10 comes from 4+6. You can do this for as many rows as you want. But what I want to point out is that if you call the first row, row zero, and the second row to be row one, and the 3rd row to be row two, and so on you will notice that in row 2 the coefficients are 1, 2, 1 for 1*a^2 +2ab + b^2 and in row 3 the coefficients are 1, 3, 3, 1 for 1*a^3+3a^2b + 3ab^2 +1*b^3. Now let's do (a+b)^5 which would be row 5(actually the 6th):
a^5 +5a^4b +10a^3b^2 +10a^2b^3 +5ab^4 +b^5. There are a couple things to note:
One: if you add up the powers on each term they will add up to 5. Two: one variable steps up while the other steps down, this will always be true. There are more complicated cases such as (a-b)^n instead of (a+b)^n and even things like (3x+2y)^n can still be handled but this is where the binomial theorem becomes a godsend. A simple case will be
(2x+3y)^3 = (1)(2x)^3 +(3)(2x)^2(3y) +(3)(2x)(3y)^2 + (3y)^3
= 8x^3 +(3)4x^2*3y + (3)2x*9y^2 + 27y^3
= 8x^3 +36x^2*y + 54x*y^2 + 27y^3
Try it by multiplying (2x+3y)(2x+3y)(2x+3y)
Now the binomial theorem says that you can find these coefficients by taking (nCr)= n things r at a time
nCr = n(n-1)(n-2)...(n-(n-r))!/(n-r)!r! = n(n-1)(n-2)...r!/(n-r)!r! = n(n-1)(n-2).../(n-r)!
A simple case is 4C2 = 4*3*2!/2!2! = 4*3*/2*1 = 6. Anyway not to get lost but the coefficients for (a+b)^3 are 3C3, 3C2, 3C1, 3C0 and for (a+b)^5 they are 5C5, 5C4, 5C3, 5C2, 5C1, 5C0.

Math Prof