Question

已知数列an=1/n(n+2),求前n项和sn的值

1是分子

Answer 1

Sn = 1/2 ×(1/n - 1/(n+2) + 1/(n-1) - 1/(n+1) + 1/返卖正(n-2) - 1/n + ....+ 1/漏悔4 -1/6 + 1/配早3 - 1/5 + 1/2 - 1/4 + 1 - 1/3) = 3/4 - 1/2(n+2) - 1/2(n+1)
这种方法叫做裂项相消

Answer 2

an=1/2(1/n-1/拆告n+2),所以sn=1/裤御携2 (1-1/3+1/3+-1/5+1/5-1/7+.....+1/n-1/胡伏n+2)=1/2(1-1/n+2)=(n+1)/(2n+4)

Answer 3

裂项 sn=1/1*2+1/2*3+……+1/n(n+2)=1-1/2+1/做段绝2-1/3+……+1/n-1/（n+1）纯姿=n/（燃冲n+1）