设数列{An}满足An+1=An^2-nAn+1,n为正整数,当A1>=3时,证明对所有的n>=1,有（1）an>=n+2(2)1/(1+a1) + 1/(1+a2) + ……+1/(1+an) =< 1/2同志们谢谢了特别是第二问,步骤谢谢万分感谢

设数列{an}满足an=2an-1+n 若{an}是等差数列,求{an}通项公式 数列{an}满足a1=1 an+1=2n+1an/an+2n 设数列{an}满足lg(1+a1+a2+...+an)=n+1,求通项公式an 设数列｛an}满足a1=2,an+1=an+1/an(n=1,2,3.),证明：an>根号下（2n+1).急用 数列｛an｝满足a1=1,且an=an-1+3n-2,求an 已知数列{an}满足an+1=2an+3.5^n,a1=6.求an 数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an 设b＞0,数列an满足a1=b,an=nban-1/an-1+n-1（n≥2）求数列an通向公式. 设b＞0,数列an满足a1=b,an=nban-1/an-1+n-1（n≥2）求数列an通向公式 已知数列An满足An＞0,其前n项和为Sn为满足2Sn=An的平方+An(1)求An(2)设数列Bn满足An/2的n次方,Tn=b1+b2+ 设数列AN满足A1等于1,3(A1+a2+~+AN)=(n+2)an,求通向公式 设数列{an}中,a1=2,an+1=an+n+1,则通项an=? 设数列{an}中,若an+1 =an+ an+2 (n∈N*),则称数列{an}为“凸数列” .设数列{an}为“凸数列”求第二问证明设数列{an}中,若an+1 =an+ an+2 (n∈N*),则称数列{an}为“凸数列” .设数列{an}为“凸数列”,若a1 =1, 设数列an满足a1=2,a(n+1)=3an+2^(n-1),求an2,设数列an满足a1=2,a(n+1)=3an+2n,求an 数列{an}满足a1=3/2,an+1=an2-an+1,求证：1/an=1/(an)-1 - 1/(an+1)-1数列{an}满足a1=3/2,an+1=an2-an+1,求证：1/an=1/(an)-1 - 1/(an+1)-1设Sn=1/a1+1/a2+...+1/an,n>2证明1 已知数列{an}满足：a1=3,an+1=（3an-2）/an ,n∈N*．（Ⅰ）证明数列{(an-1)/an-2已知数列{an}满足：a1=3,an+1=（3an-2）/an ,n∈N*．（1）证明数列{(an-1)/an-2 }为等比数列,并求数列{an}的通项公式；（2）设设b 已知数列{an}满足a1=-1,an=[(3n+3)an+4n+6]/n,bn=3^(n-1)/an+2.求数列an的通向公式.设数列bn是的前n项和已知数列{an}满足a1=-1,an=[(3n+3)an+4n+6]/n,bn=3^(n-1)/an+2.(1)求数列an的通向公式.(2)设数列bn是的前n项和为sn, 已知以1为首项数列{an}满足： an +1（n为奇数） an+1={an/2（n为偶数）}设数列{an}前n项和为sn,求数列{sn}前n项和Tn