# Summation: Prove that for every integer n > 1,...?

Prove that for every integer n > 1, ∑ (k+1)*(-1)^(k)*(n choose k+1)=0, (from k=0 to n-1).

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### 1 Answer

- MewtwoLv 58 years agoFavourite answer
Let's reduce the summand:

(k+1)*(-1)^(k)*C(n,k+1) = (k+1)*(-1)^(k)*n! / [(k+1)! (n - k - 1)!]

= n! (-1)^k / [k! (n - k - 1)!]

= n (-1)^k (n - 1)! / [k! (n - k - 1)!]

= n (-1)^k C(n-1,k).

Now, as k varies discretely from 0 to n - 1, you have that C(n -1, 0) = C(n -1, n - 1) = 1, C(n - 1, 1) = C(n - 1, n - 2) = n -1, C(n - 1, 2) = C(n - 1, n - 3) = n (n-1), and so on and so forth. But, each of these has opposite signs because of the (-1)^k. Because the sum is finite, it must converge, and we may rearrange the terms. So, ∑ (k+1)*(-1)^(k) C(n, k + 1) = n ∑ (k+1)*(-1)^(k) C(n - 1, k) = n[ (C(n - 1, 0) - C(n - 1, n - 1)) + (C(n - 1,1) - C(n - 1, n - 2)) + ... ] = n [0] = 0, as desired.

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