Question

1*2+2*3+3*4+…+n(n+1)=?请详细一点，要过程和结果，1^2+2^2+3^2+…+n^2怎么算啊，谢谢！

Answer 1

n³-(n-1)³=3n²-3n+1
(n-1)³-(n-2)³=3(n-1)²-3(n-1)+1
…
2³-1³=3*1²-3*1+1
相加
n³-1=3*(1²+…+n²)-3(1+…+n)+n
所以1²+…+n²=n(n+1)(2n+1)/6
原式=n(n+1)(2n+1)/6+n(n+1)/2=n(n+1)(n+2)/3

Answer 2

设S=1*2+2*3+……n*(n+1)
则3S=1*2*3+2*3*3+……n*(n+1)*3
=1*2(3-0)+2*3(4-1)+3*4(5-2)+……n(n+1)[n+2-(n-1)]
=1*2*3+2*3*4-1*2*3+……+n(n+1)(n+2)-(n-1)n(n+1)
=n(n+1)(n+2)，
S=n(n+1)(n+2)/3

Answer 3

此题不需要计算平方和，用下面的裂项法：

原式=(1/3)*(1*2*(3-0)+2*3*(4-1)+...+n(n+1)((n+2)-(n-1)))
=(1/3)*(1*2*3-0*1*2+2*3*4-1*2*3+...n(n+1)(n+2)-(n-1)n(n+1))
=n(n+1)(n+2)/3

Answer 4

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