Question

1/1×2×3,1/2×3×4,1/3×4×5....求数列前n项和

Answer 1

解：
由题意得：
an
=1/n×(n+1)×(n+2)
=(1/2)*[1/n×(n+1)-1/(n+1)×(n+2))]
这里能够看出：先提取一个1/2，再变成两个数相减的形式

所以整个求和数列先提取一个1/2
再把内部拆开的两个相减的数全部相加起来
可以消去中间的所有项
只留下首位项

Sn
=1/1×2×3+1/2×3×4+1/3×4×5+.........+1/n×(n+1)×(n+2)
=(1/2)*[(1/1×2-1/2×3)+(1/2×3-1/3×4)+(1/3×4-1/4×5)+........+(1/n×(n+1)-1/(n+1)×(n+2))]
=(1/2)*[1/2-1/6+1/6-1/12+1/12-1/20+...........-1/n×(n+1)+1/n×(n+1)-1/(n+1)×(n+2)]
=(1/2)*[1/2-1/(n+1)×(n+2)]
=n(n+3)/4(n+1)(n+2)