Question

1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+..........+1/(9*10*11)=?

Answer 1

1/(k-1)k(k+1)
=1/k(k^2-1)
=k/(k^2-1)-1/k
=1/2(k-1)+1/2(k+1)-1/k
=[1/(k-1)+1/(k+1)]/2-1/k

就按这种思路把每项分解开再求和。

原式==[1/1+1/3]/2-1/2
+[1/2+1/4]/2-1/3
+[1/3+1/5]/2-1/4
+[1/4+1/6]/2-1/5
...
+[1/9+1/11]/2-1/10
上式各项同乘以2，整体再除以2，其值不变。

∴原式
={[1/1+1/3]-2/2
+[1/2+1/4]-2/3
+[1/3+1/5]-2/4
+[1/4+1/6]-2/5
...
+[1/9+1/11]-2/10}/2

={1/2-1/10+1/11}/2
= 27/110

Answer 2

=1/2*(1/2-1/2*3+1/2*3-1/3*4+...+1/9*10-1/10*11)
=1/2*(1/2-1/110)
=27/110