数列综合题已知函数f（x）=ax
(1)
f(x)定义域(0,＋∞)
f(1)=a-b-0=0,a=b
f'(x)=a+a/x^2-2/x
令t=1/x,t∈(0,+∞)
f'(x)=at^2-2t+a
f(x)在其定义域内为单调函数,f'(x)>=0或f'(x)0,△=(-2)^2-4a^2=1,a>=1
若a=0,所以要满足f'(x)=1
(2)
f'(1)=a+a-2=0,a=1
a(n+1)=1+[a(n)-n+1]^2-2[(a(n)-n+1]-n^2+1
=a(n)^2-2na(n)+1
i)a1=4=2*1+2
ii)假设a(k)>=2k+2 (k>=1)
a(k+1)=a(k)^2-2ka(k)+1
=[a...全部
(1)
f(x)定义域(0,＋∞)
f(1)=a-b-0=0,a=b
f'(x)=a+a/x^2-2/x
令t=1/x,t∈(0,+∞)
f'(x)=at^2-2t+a
f(x)在其定义域内为单调函数,f'(x)>=0或f'(x)0,△=(-2)^2-4a^2=1,a>=1
若a=0,所以要满足f'(x)=1
(2)
f'(1)=a+a-2=0,a=1
a(n+1)=1+[a(n)-n+1]^2-2[(a(n)-n+1]-n^2+1
=a(n)^2-2na(n)+1
i)a1=4=2*1+2
ii)假设a(k)>=2k+2 (k>=1)
a(k+1)=a(k)^2-2ka(k)+1
=[a(k)-k]^2-k^2+1
>=[2k+2-k]^2-k^2+1
=4k+5>2(k+1)+2
iii)由i)ii)得n∈N*,a(n)>=2n+2(仅当n=1时取等号)
(3)
1+a(n+1)=a(n)^2-2na(n)+2=a(n)*[a(n)-2n]+2>=2[1+a(n)]
>=2^2[1+a(n-1)]>=。
。。。>=2^n[1+a1]=5*2^n
1/[1+a(n)]=2)
[1/（1+a1）]+[1/（1+a2）]+。。。+[1/(1+an)]
<=1/5+1/5*1/2+1/5*1/2^2+。
。。+1/5*1/2^(n-1)
=1/5[1+1/2+1/2^2+。。。+1/2^(n-1)]
=1/5[2-1/2^(n-1)]<2/5。收起
