Te art of mathematics

Algebra Level 1

∑ k = 0 n ( n k ) = ? \large \sum_{k=0}^n \dbinom{n}{k} = \ ?

2 n 2^n 2 − n 2^{-n} ( p − n ) ! (p-n)! ( n − p ) ! (n-p)!

Reda Boushab by Reda Boushab , 25, Morocco

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1 solution

Kay Xspre
Oct 1, 2015

The most simple solution is to set ( x + 1 ) n (x+1)^n and expand it with binomial theorem. It will be ( x + 1 ) n = ( n 0 ) x n + ( n 1 ) x n − 1 + ( n 2 ) x n − 2 + ⋯ + ( n n − 2 ) x 2 + ( n n − 1 ) x 1 + ( n n ) (x+1)^n = \dbinom{n}{0}x^n+ \dbinom{n}{1}x^{n-1}+\dbinom{n}{2}x^{n-2}+\dots+\dbinom{n}{n-2}x^{2}+\dbinom{n}{n-1}x^{1}+\dbinom{n}{n}

We just substitute x = 1 x = 1 , then we will get 2 n = ∑ k = 0 n ( n k ) 2^n = \sum_{k=0}^n \dbinom{n}{k}

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