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)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 onSome points to note - The first column is a constant.
- The second column is an arithmetic sequence whose common difference is 1. i.e. the terms are represented by n/1.
- 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 an^{2}+ 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)(n^{2}+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^{n}C_{2}. - Following the above process and assuming the sequence depends on third power and solving and replacing n by (n-2) we get
^{n}C_{3}. - Following the above process and assuming the sequence depends on fourth power and solving and replacing n by (n-3) we get
^{n}C_{4}. - Following the above process we get
^{n}C_{5}. - Following the above process we get
^{n}C_{6}.
^{th} power isThe binomial theorem(a + b)^{n} = ^{n}C_{0}a^{n} + ^{n}C_{1}a^{n-1}b + ^{n}C_{2}a^{n-2}b^{2} + ... + ^{n}C_{n-1}ab^{n-1} + ^{n}C_{n}b^{n} |

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