重点第三小题 已知在数列{an}中,a1=1,nan+1=2(a1+a2+a3+...+an)(n∈N*)(1)求a2,a3,a4(2)求an的通项公式(3)设数列{bn}满足b1=1/2,b(n+1)=1/ak*bn^2+bn,求证bn
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重点第三小题 已知在数列{an}中,a1=1,nan+1=2(a1+a2+a3+...+an)(n∈N*)(1)求a2,a3,a4(2)求an的通项公式
(3)设数列{bn}满足b1=1/2,b(n+1)=1/ak*bn^2+bn,求证bn
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(1)由题,nan+1=2Sn,a1=1
a2=2S1=2a1=2
a3=1/2*2S2=S2=a1+a2=3
a4=1/3*2S3=2/3[a1+a2+a3]=4
(2)nan+1=n(Sn+1 - Sn)=2Sn
Sn+1/Sn=(n+2)/n
Sn/Sn-1=(n+1)/(n-1) (n≥2)
累积得
Sn/S1=Sn/a1=Sn=(n+1)/(n-1)*n/(n-2)*(n-1)/(n-3)*...*3/1
=1/2n(n+1) (n≥2)
故an+1=2Sn/n=n+1 (n≥2)
代入n=1,2也满足,故an=n (n∈N*)
(3) n≤k,则ak=k≥n,
b(n+1)=1/ak bn^2+bn =1/k bn^2+bn
bn+1 - bn=1/k bn^2>0 ,故 {bn}数列单增
bn+1=1/k bn^2+bn=bn(bn+k)/k
求倒数
1/bn+1=1/bn - 1/(bn +k)
1/bn-1/bn+1=1/(bn+k)
nan+1=2(a1+a2+a3+...+an)(n∈N*)
n = 1 得 a2 = 2 a1 = 2
n = 2 得 2a3 = 2 ( a1 + a2 ) = 2 ( 1 + 2) 得 a3 = 3
n = 3 得 3a4 = 2 ( a1 + a2 + a3 ) = 2 ( 1 + 2 +3 ) = 12 得 a4 = 4
an的通项公式 an = n;
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nan+1=2(a1+a2+a3+...+an)(n∈N*)
n = 1 得 a2 = 2 a1 = 2
n = 2 得 2a3 = 2 ( a1 + a2 ) = 2 ( 1 + 2) 得 a3 = 3
n = 3 得 3a4 = 2 ( a1 + a2 + a3 ) = 2 ( 1 + 2 +3 ) = 12 得 a4 = 4
an的通项公式 an = n;
(3) b1=1/2,b(n+1)=1/k*bn^2+bn
bn^2+kbn < k 得 (bn+k/2)^2 < k + (k^2)/4 即 bn+k/2 < 根号下(k + (k^2)/4)
即 bn < 根号下(k + (k^2)/4) - k/2
当 k = 1时很明显成立
当 k > 1时 用归纳法证明 当 n=1时 bn < 根号下(k + (k^2)/4) - k/2成立
当n > 1 小于k时(因为bn < 根号下(k + (k^2)/4) - k/2 就是b(n+1) < 1 )
证明bn < 根号下(k + (k^2)/4) - k/2即可
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