1的平方+2的平方+3的平方+…+n的平方=1/6*n(n+1)(2n+1),求数列1*2,2*3,3*4,…,n(n+1)

问题描述:

1的平方+2的平方+3的平方+…+n的平方=1/6*n(n+1)(2n+1),求数列1*2,2*3,3*4,…,n(n+1)

因为n(n+1)=(n+1)-(n+1) 1*2+2*3+3*4+…+n(n+1) =2-2+3-3+4-4+…+(n+1)-(n+1) =2+3+4+…(n+1)-[2+3+4+…+(n+1)] =1+2+3+4+…(n+1)-[1+2+3+4+…+(n+1)] =1/6*(n+1)(n+2)(2n+3)-(n+1)(n+2)/2 =1/6*(n+1)(n+2)[(2n+3)-3] =1/3*n(n+1)(n+2)