组合数学 —— 常用组合公式

∑nk=1(2k−1)2=n(4n2−1)3 ∑ k = 1 n ( 2 k − 1 ) 2 = n ( 4 n 2 − 1 ) 3

∑nk=1k3=(n(n+1)2)2 ∑ k = 1 n k 3 = ( n ( n + 1 ) 2 ) 2

∑nk=1(2k−1)3=n2(2n2−1) ∑ k = 1 n ( 2 k − 1 ) 3 = n 2 ( 2 n 2 − 1 )

∑nk=1k4=n(n+1)(2n+1)(3n2+3n+1)30 ∑ k = 1 n k 4 = n ( n + 1 ) ( 2 n + 1 ) ( 3 n 2 + 3 n + 1 ) 30

∑nk=1k5=n2(n+1)2(2n2+2n−1)12 ∑ k = 1 n k 5 = n 2 ( n + 1 ) 2 ( 2 n 2 + 2 n − 1 ) 12

∑nk=1k(k+1)=n(n+1)(n+2)3 ∑ k = 1 n k ( k + 1 ) = n ( n + 1 ) ( n + 2 ) 3

∑nk=1k(k+1)(k+2)=n(n+1)(n+2)(n+3)4 ∑ k = 1 n k ( k + 1 ) ( k + 2 ) = n ( n + 1 ) ( n + 2 ) ( n + 3 ) 4

∑nk=1k(k+1)(k+2)(k+3)=n(n+1)(n+2)(n+3)(n+4)5 ∑ k = 1 n k ( k + 1 ) ( k + 2 ) ( k + 3 ) = n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) 5