1/1*2*3+1/2*3*4+……1/n(n+1)(n+2)=?
答案:1 悬赏:40 手机版
解决时间 2021-01-31 01:19
- 提问者网友：锁深秋
- 2021-01-30 17:30
1/1*2*3+1/2*3*4+……1/n(n+1)(n+2)=?
最佳答案
- 五星知识达人网友：鱼芗
- 2021-01-30 17:40
1/1*2*3+1/2*3*4+……1/n(n+1)(n+2)
=1/2(1/1*2-1/2*3)+1/2(1/2*3-1/3*4)+...+1/2[1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/1*2-1/2*3+1/2*3-1/3*4+...+1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/1*2-1/(n+1)(n+2)]
=1/2*[(n+1)(n+2)-2]/2(n+1)(n+2)
=(n^2+3n)/4(n+1)(n+2)
=n(n+3)/[4(n+1)(n+2)]
=1/2(1/1*2-1/2*3)+1/2(1/2*3-1/3*4)+...+1/2[1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/1*2-1/2*3+1/2*3-1/3*4+...+1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/1*2-1/(n+1)(n+2)]
=1/2*[(n+1)(n+2)-2]/2(n+1)(n+2)
=(n^2+3n)/4(n+1)(n+2)
=n(n+3)/[4(n+1)(n+2)]
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