Newton And His Binomial

Probability Level 2

( n 0 ) − ( n 1 ) + ( n 2 ) − … + ( − 1 ) n ( n n ) \displaystyle{\binom{n}{0}-\binom{n}{1}+\binom{n}{2} -\,\ldots\, + \left(-1\right)^n\binom{n}{n}}

Which of the following values is equal to the expression above?

2 n 2^n 0 0 − 1 -1 None of the other answers 1 1

Ervin Tronati by Ervin Tronati , 22, Italy

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1 solution

Sabhrant Sachan
Jun 2, 2016

Relevant wiki: Binomial Theorem

( 1 + x ) n = ( n 0 ) + ( n 1 ) x + ( n 2 ) x 2 + ( n 3 ) x 3 + ⋯ + ( n n ) x n Put x = − 1 0 = ( n 0 ) − ( n 1 ) + ( n 2 ) − ( n 3 ) + ⋯ + ( n n ) ( − 1 ) n (1+x)^{n}=\dbinom{n}{0}+\dbinom{n}{1}x+\dbinom{n}{2}x^2+\dbinom{n}{3}x^3+\cdots+\dbinom{n}{n}x^n \\ \text{Put } x=-1 \\ \boxed{0}=\dbinom{n}{0}-\dbinom{n}{1}+\dbinom{n}{2}-\dbinom{n}{3}+\cdots+\dbinom{n}{n}(-1)^n

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