题目内容
5.设各项均为正数的数列{an}的前n项和是Sn,已知a1=3,4Sn=an2+2an+4(n≥2).(1)求a2,a3;
(2)求{an}的通项公式;
(3)求证:$\frac{1}{S_1}$+$\frac{1}{S_2}$+…+$\frac{1}{S_n}$<$\frac{5}{6}$.
分析 (1)由a1=3,4Sn=an2+2an+4(n≥2),令n=2,3代入求a2,a3;
(2)由4Sn=an2+2an+4可推出4an=(an2+2an+4)-(an-12+2an-1+4),从而可得an=an-1+2,从而求{an}的通项公式;
(3)由题意可求得Sn=n2+n+1;再由$\frac{1}{{s}_{n}}$=$\frac{1}{{n}^{2}+n+1}$<$\frac{1}{{n}^{2}+n}$=$\frac{1}{n}$-$\frac{1}{n+1}$;从而证明即可.
解答 解:(1)∵a1=3,4S2=a22+2a2+4,
∴4(3+a2)=a22+2a2+4,
∴a2=4或a2=-2(舍去);
∴4(3+4+a3)=a32+2a3+4,
∴a3=6或a3=-4(舍去);
(2)当n≥3时,4Sn=an2+2an+4,①
4Sn-1=an-12+2an-1+4②
①-②得,
4an=(an2+2an+4)-(an-12+2an-1+4);
即(an+an-1)(an-2-an-1)=0;
∴an=an-1+2,
综上所述,an=$\left\{\begin{array}{l}{3,n=1}\\{2n,n≥2}\end{array}\right.$;
(3)证明:当n=1时,S1=a1=3,
当n≥2时,Sn=3+$\frac{4+2n}{2}$(n-1)=n2+n+1;
当n=1时,Sn=n2+n+1也成立;
故Sn=n2+n+1;
故$\frac{1}{{s}_{n}}$=$\frac{1}{{n}^{2}+n+1}$<$\frac{1}{{n}^{2}+n}$=$\frac{1}{n}$-$\frac{1}{n+1}$;
故$\frac{1}{S_1}$+$\frac{1}{S_2}$+…+$\frac{1}{S_n}$=$\frac{1}{3}$+$\frac{1}{7}$+$\frac{1}{13}$+…+$\frac{1}{{n}^{2}+n+1}$
=$\frac{1}{3}$+($\frac{1}{2}$-$\frac{1}{3}$)+($\frac{1}{3}$-$\frac{1}{4}$)+…+($\frac{1}{n}$-$\frac{1}{n+1}$)
=$\frac{1}{3}$+$\frac{1}{2}$-$\frac{1}{n+1}$=$\frac{5}{6}$-$\frac{1}{n+1}$<$\frac{5}{6}$.
点评 本题考查了数列的通项公式的求法及前n项和的求法,同时考查了数列与不等式的应用,属于中档题.
A. | 1 | B. | $\sqrt{2}$-1 | C. | 0.5 | D. | 0.4 |
A. | 2008 | B. | 2014 | C. | 2012 | D. | 2013 |