题目内容
5.已知等差数列{an}的前n项和为Sn,公比为q的等比数列{bn}的首项$\frac{1}{2}$,且a1+2q=3,a2+4b2=6,S5=40.(1)求数列{an},{bn}的通项公式an,bn;
(2)求数列{$\frac{1}{{a}_{n}{a}_{n+1}}$+$\frac{1}{{b}_{n}{b}_{n+1}}$}的前n项和Tn.
分析 (1)运用等差数列的通项和求和公式及等比数列的通项,列方程,解得即可得到所求通项;
(2)化简所求数列,结合裂项相消求和和等比数列的求和公式,化简整理即可得到.
解答 解:(1)设等差数列{an}的公差为d,
则$\left\{\begin{array}{l}{{a}_{1}+2q=3}\\{{a}_{1}+d+2q=6}\\{5{a}_{1}+\frac{5×4}{2}d=40}\end{array}\right.$,解得$\left\{\begin{array}{l}{{a}_{1}=2}\\{d=3}\\{q=\frac{1}{2}}\end{array}\right.$,
所以an=2+3(n-1)=3n-1,
bn=$\frac{1}{2}$•($\frac{1}{2}$)n-1=($\frac{1}{2}$)n;
(2)$\frac{1}{{a}_{n}{a}_{n+1}}$+$\frac{1}{{b}_{n}{b}_{n+1}}$=$\frac{1}{3}$($\frac{1}{{a}_{n}}$-$\frac{1}{{a}_{n+1}}$)+$\frac{1}{{b}_{n}{b}_{n+1}}$=$\frac{1}{3}$($\frac{1}{3n-1}$-$\frac{1}{3n+2}$)+22n+1,
即有Tn=$\frac{1}{3}$[($\frac{1}{2}-\frac{1}{5}$)+($\frac{1}{5}-\frac{1}{8}$)+…+($\frac{1}{3n-1}$-$\frac{1}{3n+2}$)]+$\frac{8(1-{4}^{n})}{1-4}$
=$\frac{1}{3}$($\frac{1}{2}$-$\frac{1}{3n+2}$)+$\frac{1}{3}$(22n+3-8)
=$\frac{1}{3}$(22n+3-$\frac{1}{3n+2}$)-$\frac{5}{2}$.
点评 本题考查等差数列和等比数列的通项和求和公式的运用,同时考查裂项相消求和,考查化简整理的运算能力,属于中档题.
A. | 2 | B. | 3 | C. | $\frac{4}{15}$ | D. | $\frac{4}{5}$ |