设数列{an}满足：an+1=an^2-nan+1,n=1,2,3,…当a1≥3时,证明对所有的n≥1,有(1)an≥n+2,(2)1/(1+a设数列{an}满足：an+1=an^2-nan+1,n=1,2,3,…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+a3)+……+1/

设数列{an}满足：an+1=an^2-nan+1,n=1,2,3,…当a1≥3时,证明对所有的n≥1,有(1)an≥n+2,(2)1/(1+a设数列{an}满足：an+1=an^2-nan+1,n=1,2,3,…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+a3)+……+1/
设数列{an}满足：an+1=an^2-nan+1,n=1,2,3,…当a1≥3时,证明对所有的n≥1,有(1)an≥n+2,(2)1/(1+a

设数列{an}满足：an+1=an^2-nan+1,n=1,2,3,…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+a3)+……+1/(1+an)≤1/2

设数列{an}满足：an+1=an^2-nan+1,n=1,2,3,…当a1≥3时,证明对所有的n≥1,有(1)an≥n+2,(2)1/(1+a设数列{an}满足：an+1=an^2-nan+1,n=1,2,3,…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+a3)+……+1/
用数学归纳法：当n=1时显然成立,假设当n≥k时成立即
ak≥k+2,则当n=k+1时,ak+1=ak(ak-k)+1≥ak(k+2-k)+1≥(k+2)·2+1＞k+3,成立.
（2）利用上述部分放缩的结论ak+1≥2ak+1来放缩通项,可得ak+1+1≥2(ak+1) ak+1≥…
≥2k-1(a1+1)≥2k-1·4=2k+1≤.