题目内容
13.设数列{an}满足${a_1}+3{a_2}+{3^2}{a_3}+…+{3^{n-1}}a{\;}_n=\frac{n}{3}$,n∈N*.(1)a1,a2;
(2)求数列{an}的通项公式;
(3)设bn=$\frac{1}{{{{log}_3}{a_n}•{{log}_3}{a_{n+1}}}}$,求{bn}的前n项和Sn.
分析 (1)由题意得a1=$\frac{1}{3}$,a1+3a2=$\frac{2}{3}$;从而求a1,a2;
(2)当n≥2时,由${a_1}+3{a_2}+{3^2}{a_3}+…+{3^{n-1}}a{\;}_n=\frac{n}{3}$可得3n-1an=$\frac{1}{3}$;从而解得;
(3)化简bn=$\frac{1}{{{{log}_3}{a_n}•{{log}_3}{a_{n+1}}}}$=$\frac{1}{(-n)(-(n+1))}$=$\frac{1}{n}$-$\frac{1}{n+1}$,从而由{bn}的前n项和Sn.
解答 解:(1)由题意得,
a1=$\frac{1}{3}$,a1+3a2=$\frac{2}{3}$;
解得,${a_1}=\frac{1}{3},{a_2}=\frac{1}{9}$;
(2)当n≥2时,
${a_1}+3{a_2}+{3^2}{a_3}+…+{3^{n-1}}a{\;}_n=\frac{n}{3}$,①
a1+3a2+32a3+…+3n-2an-1=$\frac{n-1}{3}$,②
①-②得,
3n-1an=$\frac{1}{3}$;
解得:${a_n}={(\frac{1}{3})^n}$;
a1=$\frac{1}{3}$也成立;
故${a_n}={(\frac{1}{3})^n}$.
(3)∵bn=$\frac{1}{{{{log}_3}{a_n}•{{log}_3}{a_{n+1}}}}$=$\frac{1}{(-n)(-(n+1))}$=$\frac{1}{n}$-$\frac{1}{n+1}$,
∴Sn=(1-$\frac{1}{2}$)+($\frac{1}{2}$-$\frac{1}{3}$)+…+($\frac{1}{n}$-$\frac{1}{n+1}$)
=1-$\frac{1}{n+1}$=$\frac{n}{n+1}$.
点评 本题考查等比数列的求法及裂项求和法的应用,属于基础题.
x | -1 | 0 | 1 | 2 | 3 |
f(x) | -0.677 | 3.011 | 5.432 | 5.980 | 7.651 |
g(x) | -0.530 | 3.451 | 4.890 | 5.241 | 6.892 |
A. | (-1,0) | B. | (1,2) | C. | (0,1) | D. | (2,3) |
A. | $\frac{3}{5}$ | B. | $\frac{1}{3}$ | C. | $\frac{\sqrt{2}}{2}$ | D. | $\frac{\sqrt{3}}{2}$ |
A. | 判断 | B. | 有向线 | C. | 循环 | D. | 开始 |
A. | 0 | B. | 1 | C. | 2 | D. | 3 |