高二数列题:设数列{an}满足an+1=an^2-nan+1,n为正整数,当a1>=3时,证明……设数列{an}满足an+1=an^2-nan+1,n为正整数,当a1>=3时,证明(1)an>=n+2(2)1/(1+a1) + 1/(1+a2) + ……+1/(1+an) =< 1/2
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![高二数列题:设数列{an}满足an+1=an^2-nan+1,n为正整数,当a1>=3时,证明……设数列{an}满足an+1=an^2-nan+1,n为正整数,当a1>=3时,证明(1)an>=n+2(2)1/(1+a1) + 1/(1+a2) + ……+1/(1+an) =< 1/2](/uploads/image/z/5466758-14-8.jpg?t=%E9%AB%98%E4%BA%8C%E6%95%B0%E5%88%97%E9%A2%98%EF%BC%9A%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3an%2B1%3Dan%5E2-nan%2B1%2Cn%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E5%BD%93a1%3E%3D3%E6%97%B6%2C%E8%AF%81%E6%98%8E%E2%80%A6%E2%80%A6%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3an%2B1%3Dan%5E2-nan%2B1%2Cn%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E5%BD%93a1%3E%3D3%E6%97%B6%2C%E8%AF%81%E6%98%8E%EF%BC%881%EF%BC%89an%3E%3Dn%2B2%282%291%2F%281%2Ba1%29+%2B+1%2F%281%2Ba2%29+%2B+%E2%80%A6%E2%80%A6%2B1%2F%281%2Ban%29+%3D%3C+1%2F2)
高二数列题:设数列{an}满足an+1=an^2-nan+1,n为正整数,当a1>=3时,证明……设数列{an}满足an+1=an^2-nan+1,n为正整数,当a1>=3时,证明(1)an>=n+2(2)1/(1+a1) + 1/(1+a2) + ……+1/(1+an) =< 1/2
高二数列题:设数列{an}满足an+1=an^2-nan+1,n为正整数,当a1>=3时,证明……
设数列{an}满足an+1=an^2-nan+1,n为正整数,当a1>=3时,证明
(1)an>=n+2
(2)1/(1+a1) + 1/(1+a2) + ……+1/(1+an) =< 1/2
高二数列题:设数列{an}满足an+1=an^2-nan+1,n为正整数,当a1>=3时,证明……设数列{an}满足an+1=an^2-nan+1,n为正整数,当a1>=3时,证明(1)an>=n+2(2)1/(1+a1) + 1/(1+a2) + ……+1/(1+an) =< 1/2
(1)当n=1时,a1>=3=1+2,an>=n+2成立;
当n >1时,an=(an-1)^2-nan-1+1,令S= an-(n+2)
=(an-1)^2-nan-1+1-(n+2)
=(an-1)^2-(n+1)an-1-1.
我们把S看作是以an-1为变数的二次函数,其中,
二次项系数=1大于0,
△=[-(n+1)]^2-4*1*(-1)= (n+1)^2+4>0,因此,S>0,
即an>n+2成立.
结合a1>=3有,an>=n+2成立.
1)用数学归纳法。
A(n+1)=An^2-nAn+1=An(An-n)+1>=An*2+1>=(n+2)*2+1=2n+5>n+1+2
(2)因为an>=n+2,所以an-n>=2
A(n+1)=An(An-n)+1>=2An+1
A(n+1)+1>=2(An+1)
1/(A(n+1)+1)<=1/(An+1)*1/2
1/(1+a1)=1/4,1...
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1)用数学归纳法。
A(n+1)=An^2-nAn+1=An(An-n)+1>=An*2+1>=(n+2)*2+1=2n+5>n+1+2
(2)因为an>=n+2,所以an-n>=2
A(n+1)=An(An-n)+1>=2An+1
A(n+1)+1>=2(An+1)
1/(A(n+1)+1)<=1/(An+1)*1/2
1/(1+a1)=1/4,1/(1+a2)<=1/2*1/4
1/(1+a1) + 1/(1+a2) + ……+1/(1+an) <=1/4(1+1/2+1/4+.....+1/2^(n-1))<1/4*2=1/2
当且仅当n=1时取等号。
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