Sn=1/1*2+1/2*3,...,1/n*(n+1)
=（1-1/2）+（1/2-1/3）+.+[1/n-1/(n+1)]
=1-1/(n+1)
=n/(n+1)
用数学归纳法证:
当k=1时:S1=1/1*2=1/2 k/(k+1)=1/2 所以Sk=k/(k+1)
假设当k=n时成立,即:Sn=n/(n+1)
那么当k=n+1时,证明S(n+1)=(n+1)/(n+2)即可
S(n+1)=1/1*2+1/2*3,...,1/n*(n+1)+1/(n+1)(n+2)
=n/(n+1)+1/(n+1)(n+2)
=n(n+2)/(n+1)(n+2)+1/(n+1)(n+2)
=(n^2+2n+1)/(n+1)(n+2)
=(n+1)^2/(n+1)(n+2)
=(n+1)/(n+2)
所以综上:Sn=n/(n+1)
o(∩_∩)o...