已知正项数列{an}的前n项和为Sn,a1=2/3,且满足2S(n+1)+2Sn=3(an+1)^2(n属于N*)1,求数列{an}的通项公式an2,求证；当n>=2时,1/(a2)^2+1/(a3)^2+.+1/(an)^2
问题描述：
已知正项数列{an}的前n项和为Sn,a1=2/3,且满足2S(n+1)+2Sn=3(an+1)^2(n属于N*)
1,求数列{an}的通项公式an
2,求证；当n>=2时,1/(a2)^2+1/(a3)^2+.+1/(an)^2
答
an=2n/3
答
a1=2/3,s1=a1=2/3,s2=a1+a2,
2(a1+a2)+2a1=3(a2)^2,a2=4/3
2(a1+a2+a3)+2(a1+a2)=3(a3)^2,a3=2
a4=8/3
....
an=2n/3
9/16 +9/36+9/64+...+9/(4n^2)=9(1/4+1/9+1/16+...+1/(n^2))/4
答
(1)an=2n/3(递推一项,两式相减得an与an+1关系）
(2)即求n>=2时,∑1／n^2