Binomial Theorem

We are interested in finding relation between many things. When (a + b) is raised to different powers then relation between previous expansion with the next expansion has a very interesting result. This post is about this and binomial theorem. It is called binomial as it has two variables in it. In this post I will show you how we can arrive at the coefficients of the Binomial Expansion. The foundation of binomial theorem lies in Pascals Triangle. Pascals triangle is an interesting topic and I will cover it in detail. A Pascals Triangle looks like this.

                              1
                            1   1
                          1   2   1
                        1   3   3   1
                      1   4   6   4   1
                    1   5   10  10  5   1
                      so on.  
A pascals triangle is generated from coefficients of the expansion of different powers raised to (a + b). Let us expand some of the powers.

(a + b)1 = (a + b)

(a + b)2 = (a2 + 2ab + b2)

(a + b)3 = (a3 + 3a2b + 3ab2 + b3)

(a + b)4 = (a4 + 4a3b + 6a2b2 + 4ab3 + b4)

(a + b)5 = (a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5)
Let us express the coefficients in other form:
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10  10  5   1
1   6   15  20  15  6  1
so on
Some points to note
  1. The first column is a constant.
  2. The second column is an arithmetic sequence whose common difference is 1. i.e. the terms are represented by n/1.
  3. The third column needs some explanation
    The sequence is 1 3 6 10 15...
    The difference between successive terms is 2 3 4 5....
    The difference of elements again is 1 1 1.
    Don't worry about the method which I am going to follow. Learn the method. If on finding successive differences you reach at a constant in 2 steps. Suppose the sequence is formed by the second power. Here, the equation is an2 + bn + c. Where n is the position of term. Let us take the first three terms of the sequence. Then we get a system of simultaneous equations.

    a + b + c = 1
    4a + 2b + c = 3
    9a + 3b + c = 6

    solving we get, a = 1/2, b = 1/2 and c = 0
    The equation is (1/2)(n2+n). In the table the sequence start from row 2 so we will replace n by (n-1).
    Then the equation becomes (1/2)((n-1)2 + (n-1)).
    Reducing into factors we get n(n-1)/2 = n(n-1)/(1x2)
    The above thing in combinations is represented as nC2.
  4. Following the above process and assuming the sequence depends on third power and solving and replacing n by (n-2) we get nC3.
  5. Following the above process and assuming the sequence depends on fourth power and solving and replacing n by (n-3) we get nC4.
  6. Following the above process we get nC5.
  7. Following the above process we get nC6.
Hence the expansion of (a+b) raised to nth power is

The binomial theorem
(a + b)n = nC0an + nC1an-1b + nC2an-2b2 + ... + nCn-1abn-1 + nCnbn

Comments