Question

1*2+2*3+3*4.+N*(N+1) 用什么公式算啊

Answer 1

解法一：（裂项相消）

因

n(n+1) = 1/3[n(n+1)(n+2) -(n-1)n(n+1)]

所以

1*2+2*3+3*4.+N*(N+1) = 1/3(1*2*3 - 0) + 1/3(2*3*4 - 1*2*3)+1/3(3*4*5 - 2*3*4) +.+ 1/3[n(n+1)(n+2) -(n-1)n(n+1)]

= 1/3[(1*2*3 - 0)+(2*3*4 - 1*2*3)+(3*4*5 - 2*3*4)+.+[n(n+1)(n+2) -(n-1)n(n+1)]]

= 1/3n(n+1)(n+2)

解法二：（分别求和）

因 n(n+1) = n^2 + n

则

1*2+2*3+3*4.+N*(N+1)

= (1^2 + 2^2 +3^3+.+n^2) +(1+2+3+...+n)

= 1/6n(n+1)(2n+1) + 1/2n(n+1)

= 1/3n(n+1)(n+2)