题目内容

1.已知正项数列{an}的前n项和为Sn
(1)若4Sn-an2-2an-1=0,求{an}的通项公式;
(2)若{an}是等比数列,公比为q(q≠1,q为正常数),数列{lgan}的前n项和为Tn,$\frac{{T}_{(k+1)n}}{{T}_{kn}}$为定值,
求a1

分析 (1)利用4Sn-an2-2an-1=0与$4{S_{n-1}}-a_{n-1}^2-2{a_{n-1}}-1=0$作差可得an-an-1=2,进而可得结论;
(2)通过设${a_n}={a_1}{q^{n-1}}$可得数列{lgan}是lga1为首项、lgq为公差的等差数列,利用等差数列的求和公式并化简可得$\left\{\begin{array}{l}{(k+1)^{2}-p{k}^{2}=0}\\{(k+1)-pk=0或{{a}_{1}}^{2}=q}\end{array}\right.$(其中p为定值),计算即得结论.

解答 解:(1)∵4Sn-an2-2an-1=0                        ①
∴$4{a_1}-a_1^2-2{a_1}-1=-a_1^2+2{a_1}-1=-{({a_1}-1)^2}=0$,
即a1=1,
由①得:当n≥2时,$4{S_{n-1}}-a_{n-1}^2-2{a_{n-1}}-1=0$        ②
①-②得:(an+an-1)(an-an-1-2)=0,
∵an>0,∴an-an-1-2=0,即an-an-1=2,
∴数列{an}是首项为1,公差为2的等差数列,
∴an=2n-1;
(2)由题设可知:${a_n}={a_1}{q^{n-1}}$,
令bn=lgan=nlgq+lga1-lgq,
∴数列{lgan}是lga1为首项、lgq为公差的等差数列,
若$\frac{{{T_{(k+1)n}}}}{{{T_{kn}}}}$为定值,令$\frac{{{T_{(k+1)n}}}}{{{T_{kn}}}}=p$(定值),
则$\frac{{(k+1)nlg{a_1}+\frac{(k+1)n[(k+1)n-1]}{2}lgq}}{{knlg{a_1}+\frac{kn(kn-1)}{2}lgq}}=p$,
即{[(k+1)2-pk2]lgq}n+[(k+1)-pk]($lg\frac{{{a}_{1}}^{2}}{q}$)lgq=0对n∈N*恒成立             (*)
∵q≠1,q>0,
∴(*)式等价于$\left\{\begin{array}{l}{(k+1)^{2}-p{k}^{2}=0}\\{(k+1)-pk=0或{{a}_{1}}^{2}=q}\end{array}\right.$,
∴$\frac{k+1}{k}=\sqrt{p}$,将其代入(k+1)-pk=0,得:p=0或p=1,
∵k∈N*,∴p>0且p≠1,∴$a_1^2=q$,
∵an>0,∴q>0,
∴${a}_{1}=\sqrt{q}$.

点评 本题考查求数列的通项及求和,考查运算求解能力,注意解题方法的积累,属于中档题.

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