裂项相消法.
an=2/[n(n+2)]=1/n-1/(n+2)
所以,
S10=(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+...+(1/10-1/12)
=1+1/2-1/11-1/12
=3/2-23/132
=175/132