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Corollaries of Binomial Theorem

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  • Difficulty Level : Easy
  • Last Updated : 28 Jun, 2022
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The expression (a+b)^n denotes (a+b)(a+b)(a+b) ... n times. This can be evaluated as the sum of the terms involving a^k b^{n-k} for k = 0 to n, where the first term can be chosen from n places, second term from (n-1) places, k^{th} term from (n-(k-1)) places and so on. This is expressed as (a+b)^n = \sum\limits_{k=0}^n ^nC_k a^{n-k} b^k . The binomial expansion using Combinatorial symbols is

(a+b)^n = ^nC_0 a^n b^0 + ^nC_1 a^{n-1} b^1 + ^nC_2 a^{n-2} b^2 .. + ^nC_{n-k} a^k b^{n-k} .. +^nC_n a^0 b^n

  • The degree of each term a^k [Tex]b^{n-k} [/Tex]in the above binomial expansion is of the order n.
  • The number of terms in the expansion is n+1.
  • ^nC_k = n!/k!(n-k)! Similarly ^nC_{n-k} = n!/(n-k)!(n-(n-k))! = n!/(n-k)!k! Hence it can be concluded that ^nC_k = ^nC_{n-k} .

Substituting a = 1 and b = x in the binomial expansion, for any positive integer n we obtain (1+x)^n = ^nC_0 + ^nC_1 x^1 + ^nC_2 x^2 ..+ ^nC_n x^n . Corollary 1:

\sum\limits_{k=0}^n ^nC_k = 2^n

for any non-negative integer n. Replacing x with 1 in the above binomial expansion, We obtain ^nC_0 + ^nC_1 + ^nC_2 .. + ^nC_n = (1+1)^n = 2^n . Corollary 2:

\sum\limits_{k=0}^n ^nC_k = 0

for any positive integer n. Replacing x with -1 in the above binomial expansion, We obtain ^nC_0 + ^nC_1 (-1) + ^nC_2 (-1)^2 .. + ^nC_n (-1)^n = (1+(-1))^n = 0 . Corollary 3: Replacing x with 2 in the above binomial expansion, we obtain ^nC_0 + ^nC_1 2 + ^nC_2 2^2 .. + ^nC_n 2^n = (1+2)^n = 3^n In general, it can be said that

\sum\limits_{k=0}^n (2^k) ^nC_k = 3^n

Additionally, one can combine corollary 1 and corollary 2 to get another result, ^nC_0 + ^nC_1 (-1) + ^nC_2 (-1)^2 .. + ^nC_n (-1)^n = (1+(-1))^n = 0 [Tex]^nC_0 + ^nC_2 + .. = ^nC_1 + ^nC_3 + … [/Tex]Sum of coefficients of even terms = Sum of coefficients of odd terms. Since \sum\limits_{k=0}^n ^nC_k = 2^n , 2(^nC_0 + ^nC_2 + ..) = 2^n [Tex]^nC_0 + ^nC_2 + .. = 2^{n-1} [/Tex]

^nC_0 + ^nC_2 + .. = ^nC_1 + ^nC_3 + .. = 2^{n-1}

Counting The coefficients of the terms in the expansion (a+b)^n correspond to the terms of the pascal’s triangle in row n.

(a+b)^011
(a+b)^1a+b1 \ 1
(a+b)^2a^2+2ab+b^21 \ 2 \ 1
(a+b)^3a^3+3a^2b+3ab^2+b^31 \ 3 \ 3 \ 1

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