Evaluate a sum with binomial coefficients: sum_(k=0)^n (-1)^k k(n;k)^2

Adison Rogers

Adison Rogers

Answered question

2022-11-06

Evaluate a sum with binomial coefficients: k = 0 n ( 1 ) k k ( n k ) 2

Answer & Explanation

Calvin Maddox

Calvin Maddox

Beginner2022-11-07Added 15 answers

First, use k ( n 1 k ) = n ( n 1 k 1 ) = n ( n 1 n k )
(1) k = 0 n ( 1 ) k k ( n k ) 2 = n k = 0 n ( 1 ) k ( n k ) ( n 1 n k )
Next compute a generating function. The sum we want is the coefficient of x n
n m , k ( 1 ) k ( n k ) ( n 1 m k ) x m = n m , k ( 1 ) k ( n k ) ( n 1 m k ) x m k x k = n k ( 1 ) k ( n k ) ( 1 + x ) n 1 x k = n ( 1 + x ) n 1 ( 1 x ) n (2) = n ( 1 x 2 ) n 1 ( 1 x )
The sum we want is the coefficient of x n in (2)
k = 0 n ( 1 ) k k ( n k ) 2 = { n ( n 1 n / 2 ) ( 1 ) n / 2 if  n  is even n ( n 1 ( n 1 ) / 2 ) ( 1 ) ( n + 1 ) / 2 if  n  is odd (3) = n ( n 1 n / 2 ) ( 1 ) n / 2

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