Question

1*0+2*1+3*2+4*3+……+n(n-1)

1*0+2*1+3*2+4*3+……+n(n-1)=?

Answer 1

裂项法:
同乘以3后: 原式=1*2*3+2*3*3+3*4*3+....+(n-1)*n*3
=1*2*3+2*3*(4-1)+3*4*(5-2)+....(n-1)n*[(n+1)-(n-2)]
=1*2*3+2*3*4-1*2*3+3*4*4-2*3*4+(n-1)n(n+1)-(n-2)(n-1)n
=(n-1)n(n+1)
再除以3, 结果是(n-1)n(n+1)/3

Answer 2

1*0+2*1+3*2+4*3+……+n(n-1)
=1²+2²+3²+----+n²-(1+2+3+-----+n)
=n(n+1)(2n+1)/6-n(n+1)/2
=n(n+1)(n-1)/3
=(n³-n)/3