已知数列{an}中，a1=1，a1+2a2+3a3+…+nan=(n+1)2an+1(n∈N)

精选知识

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×3?

Tn-3Tn=-2Tn=1+3+...+3^(n-1)-n×3?

=1×(3?-1)/(3-1) -n×3?

=[(1-2n)×3?-1]/2

Tn=[(2n-1)×3? +1]/4

3.

[a(n+1)/(n+2)]/[an/(n+1)]

=[3?/(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

其他回答

a(1)=1,

a(1)+2a(2)+...+na(n) = (n+1)a(n+1)/2, a(1) = 2a(2)/2, a(2)=1.

a(1)+2a(2)+...+na(n)+(n+1)a(n+1) = (n+2)a(n+2)/2 = (n+1)a(n+1)/2 + (n+1)a(n+1)=3(n+1)a(n+1)/2,

(n+2)a(n+2) = 3(n+1)a(n...

其他类似问题

问题1：已知数列{an}中,a1=1,a1+2a2+3a3+…+nan=((n+1)/2)a(n+1)(n∈N*)（1）求an的通项公式 （2）求数列｛n^2an｝的前n项和Tn[数学科目]

由a1 = 1,

a1 + 2a2 + 3a3 + ...+ nan = ((n + 1) / 2)a(n + 1) (*)

(*)式取n = 1 得 a2 = 1

当k ≥ 3时

[(*)式取n = k] - [(*)式取n = k - 1] 并将k替换为n 得 nan = [(n + 1)a(n + 1) - nan] / 2

整理得 a(n + 1) / an = 3n / (n + 1)

a(n + 1) = a2 * (a3 / a2) * (a4 / a3) * ...* (a(n + 1) / an)

= (3 * 2 / 3) * (3 * 3 / 4) * ...* (3 * n / (n + 1))

= 2 * 3^(n - 1) / (n + 1)

即

a1 = 1

an = (2 / n) * 3^(n - 2) 当n ≥ 2

应该是n*2an吧 否则太难了

设 bn = n*2an

b1 = 2,

bn = 4 * 3^(n - 2) n ≥ 2 是等比数列

从而 Tn = 2 + 4 * (1 - 3^(n - 1)) / (1 - 3)

= 2 * 3^(n - 1)

恰好对n = 1,2,...成立.

问题2：已知数列an满足a1+2a2+3a3+……+nan=2^n,求an

a1+2a2+3a3+……+(n-1)a(n-1)+nan=2^n

a1+2a2+3a3+……+(n-1)a(n-1)=2^(n-1)

两式相减得

nan=2^n-2^(n-1)

nan=2^(n-1)

an=2^(n-1)/n

问题3：已知数列an满足a1+2a2+3a3+...+nan=n(n+1)*(n+2),则数列an的前n项和Sn=?[数学科目]

a1+2a2+3a3+...+nan=n(n+1)*(n+2),则：

a1+2a2+3a3+...+（n-1）×an-1=n(n-1)*(n+1),两式相减：

nan=n(n+1)*(n+2)-n(n-1)*(n+1),得

an=3n+3

所以：

a1+a2+a3+...+an=3*(1+2+3+...+n)+3n=3*n(n+1)/2+3n

整理得前n项和为:a1+a2+a3+...+an=3n(n+3)/2

问题4：已知数列{an}中,若a1+2a2+3a3+…+nan=n(n+1)(n+2)则 an=[数学科目]

若a1+2a2+3a3+…+nan=n(n+1)(n+2) (1)

则a1+2a2+3a3+...+(n-1)a(n-1)=(n-1)n(n+1) (2)

nan=n(n+1)(n+2-n+1)

an=3n+3

问题5：在数列{an}中,a1=1,a1+2a2+3a3+.+nan=(n+1)(an+1)/2,求{an}的通项an+1是指a的n+1项[数学科目]

a1=1,a1+2a2+3a3+.+nan=(n+1)(a(n+1))/2,

令n=1得：a1=2a2/2,a2=1.

当n≥2时,a1+2a2+3a3+.+(n-1)a(n-1)=na(n)/2,

两式相减得：nan=(n+1)(a(n+1))/2 -na(n)/2,

3na(n)/2=(n+1)(a(n+1))/2,

a(n+1) /a(n)= 3n/(n+1)( n≥2),

所以a3/a2=3?2/3,

a4/a3=3?3/4,

a5/a4=3?4/5,

…………

a(n) /a(n-1)= 3(n-1)/n

以上各式相乘得：a(n) / a2=3^(n-2)?2/n( n≥2),

a(n)=3^(n-2)?2/n ( n≥2),

综上可知：n=1时,a(n)=1.

n≥2时,a(n)=3^(n-2)?2/n.