已知数列{an}中,an=8n/((2n-1)^2(2n+1)^2),sn为其前n项的和,归纳sn的公式

(2n+1)^2-(2n-1)^2=4n^2+4n+1-(4n^2-4n+1)=8n
An=[(2n+1)^2-(2n-1)^2]/[(2n-1)^2(2n+1)^2]
=(2n+1)^2/[(2n-1)^2(2n+1)^2]-(2n-1)^2/[(2n-1)^2(2n+1)^2]
=1/(2n-1)^2-1/(2n+1)^2
Sn=A1+A2+A3+……A(n-1)+An
=(1/1^2-1/3^2)+(1/3^2-1/5^2)+(1/5^2-1/7^2)+……+[1/(2n-3)^2-1/(2n-1)^2]+[1/(2n-1)^2-1/(2n+1)^2]
=1-[1/(2n+1)^2]