已知数列an,a1(a2)(3a3)……(nan)=n ,(n>=1).则an=

Question

如题所述

Answer 1

1.
n≥2时,
a1+2a2+3a3+...+nan=[(n+1)/2]a(n+1) (1)
a1+2a2+3a3+...+(n-1)a(n-1)=(n/2)an (2)
(1)-(2)
nan=[(n+1)/2]a(n+1)-(n/2)an
(n+1)a(n+1)=3nan
[(n+1)a(n+1)]/(nan)=3,为定值
a1×1=1×1=1,数列{nan}是以1为首项,3为公比的等比数列
nan=1×3^(n-1)=3^(n-1)
an=3^(n-1)/n
n=1时,a1=1/1=1,同样满足通项公式
数列{an}的通项公式为an=3^(n-1)/n
2.
n^2·an=n^2·[3^(n-1)/n]=n·3^(n-1)
Tn=1×1+2×3+3×3²+...+n×3^(n-1)
3Tn=1×3+2×3²+...+(n-1)×3^(n-1)+n×3ⁿ
Tn-3Tn=-2Tn=1+3+...+3^(n-1)-n×3ⁿ
=1×(3ⁿ-1)/(3-1) -n×3ⁿ
=[(1-2n)×3ⁿ-1]/2
Tn=[(2n-1)×3ⁿ +1]/4
3.
[a(n+1)/(n+2)]/[an/(n+1)]
=[3ⁿ/(n+1)(n+2)]/[3^(n-1)/n(n+1)]
=3n/(n+2)
n≥1 n/(n+2)≤1/3,当且仅当n=1时取等号
3n/(n+2)≤1,当且仅当n=1时取等号
即对数列{an/(n+1)},a1/2=a2/3,当n≥2时,{an/(n+1)}单调递减
a1/2=1/2
要不等式an≤(n+1)λ对任意正整数n恒成立,即an/(n+1)≤λ对任意正整数n恒成立,只需当an/(n+1)取最大值时不等式成立.
λ≥1/2,λ的最小值为1/2