Question

这个数学题怎么做？？

Answer 1

因为1/[n(n+1)]=[(n+1)-n]/[n(n+1)]
=(n+1)/[n(n+1)]-n/[n(n+1)]
=1/n-1/(n+1)
即：
1/(1×2) =1/1-1/2=1-1/2
1/(2×3)=1/2-1/3
1/(3×4) =1/3-1/4
……
1/(2011×2012) =1/2011-1/2012
相加得：
1/(1×2)+1/(2×3)+1/(3×4) +……+1/(2011×2012)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2011-1/2012)
=1+(-1/2+1/2)+(-1/3+1/3)+(-1/4+1/4)+……+(-1/2011+1/2011)-1/2012
=1+0+0+0+……0-1/2012
=1-1/2012
=2011/2012
有不明白之处请留言。

Answer 2

这就是∑1/(k(k+1)) k=1到2011嘛，显然用裂项相消嘛，就是∑1/(k(k+1))=∑(1/k-1/(k+1))=2011/2012嘛

Answer 3

1/(1x2)+1/(2x3) +...+1/(2011x2012)
=(1 -1/2) +(1/2-1/3)+...+(1/2011-1/2012)
=1 -1/2012
=2011/2012

Answer 4

不会……………………

追问

emmmm...

Answer 5

因为
1/(n×(n+1))=1/n-1/(n+1)
所以
1/(1×2)+1/(2×3)+1/(3×4)+......+1/(2011×2012)
=1-1/2+1/2-1/3+1/3-1/4+....+1/2011-1/2012
=1-1/2012
=2011/2012