Question

数学题:1/(1*3)+1/(2*4)+1/(3*5)+....+1/[n*(n+2)]=

Answer 1

1/(1*3)=(1-1/3)/2
1/(2*4)=(1/2-1/4)/2
.
.
.
.原式*2=
1-1/3+1/2-1/4...............1/n-1/(n+2)=1+1/2-1/3+1/3-1/4+1/4..........1/(n+1)-1/(n+2)=1+1/2-1/(n+1)-1/(n+2)=3/2-1/(n+1)-1/(n+2)
然后把让面的结果除以二化简一下就行了。

Answer 2

俊狼猎英团队为您解答：

1/(1×3)=1/2(1/1-1/3)
1/(2×4)=1/2(1/2-1/4)
1/(3×5)=1/2(1/3-1/5)
……
1/[n(n+1)]=1/2[1/n-1/(n+1)]
以上各式相加得：
原式=1/2［1+1/2-1/(n+1)-1/(n+2)］
=1/2*[3(n+1)(n+2)-2(n+2)-2(n+1)]/[2(n+1)(n+2)]
=(3n^2+5n)/[4(n+1)(n+2)]

Answer 3

列项求合法，怎么不会哦，记住方法这种题最简单啦

Answer 4

裂项法
原式=(n+1)/(2n+4)

Answer 5

1/2*[1+1/2-1/(n+1)-1/(n+2)]