应该是1/1*3,1/3*5,…,1/(2n-1)(2n+1)的前n项和Sn吧
Sn=1/1x3+1/3x5+...+1/(2n-1)(2n+1)
=（1/2）x[(3-1)/(1x3)+(5-3)/(3x5)+...+(2n+1-(2n-1))/(2n-1)(2n+1)]
=(1/2)x(1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1))
=(1/2)(1-1/(2n+1))
=1/2-1/(4n+2)
1/n(n+2)=1/2[1/n-1/(n+2)]
所以Sn=1/2[1-1/3+1/3-1/5+......+1/n-1/(n+2)]=1/2[1-1/(n+2)]=(n+1)/(2n+4)
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