作业帮1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+……1/(98*99*100)简便算法及答案
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/09 03:37:39
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+……1/(98*99*100)简便算法及答案
首先写出这个式子的通项
a(n)=1/(n*(n+1)*(n+2))
所以
a(n)+a(n+1)
=1/(n*(n+1)*(n+2))+1/((n+1)*(n+2)*(n+3))
=(2n+3)/(n*(n+1)*(n+2)*(n+3))
=1/(n*(n+2))-1/((n+1)*(n+3))
写成这个样子
就很简单了
a1+a2=1/1*3-1/2*4
a2+a3=1/2*4-1/3*5
……
所以
(a1+a2)+(a2+a3)+…+(a97+a98)
=(1/1*3-1/2*4)+(1/2*4-1/3*5)+…+(1/97*99-1/98*100)
=1/1*3-1/98*100
这个结果加上头尾两个
即a1和a98就是题目所求的两倍
（1/3-1/9800+1/6+1/9800*99）/2
=（1/2-1/9900）/2
=4959/19800
答案是 4959/19800
1/[n(n+1)(n+2)]
=2{1/[n(n+1)]-1/[(n+1)(n+2)]}
=2{[1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}
所以
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+……1/(98*99*100)
=2[(1/1-1/2)-(1/2-1/3)]+2[(1/2-1/3)-(1/3-1/4)]+……...
全部展开
1/[n(n+1)(n+2)]
=2{1/[n(n+1)]-1/[(n+1)(n+2)]}
=2{[1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}
所以
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+……1/(98*99*100)
=2[(1/1-1/2)-(1/2-1/3)]+2[(1/2-1/3)-(1/3-1/4)]+……+2[(1/98-1/99)-(1/99-1/100)]
=2{[(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/98-1/99)]
-[(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+……+(1/99-1/100)]}
=2[(1-1/99)-(1/2-1/100)]=4949/4950
收起