题目内容
9.已知递增的等比数列{an}前三项之积为8,且这三项分别加上1、2、2后又成等差数列.(1)求等比数列{an}的通项公式;
(2)记bn=an+2n,求数列{bn}的前n项和Tn.
分析 (1)利用等差数列与等比数列的通项公式即可得出;
(2)利用等差数列与等比数列的前n项和公式即可得出.
解答 解:(1)设等比数列前三项分别为a1,a2,a3,
则a1+1、a2+2、a3+2又成等差数列.
依题意得:$\left\{\begin{array}{l}{{a}_{1}{a}_{2}{a}_{3}=8}\\{2({a}_{2}+2)=({a}_{1}+1)({a}_{3}+2)}\end{array}\right.$,
即$\left\{\begin{array}{l}{{a}_{1}•{a}_{1}q•{a}_{1}{q}^{2}=8}\\{2({a}_{1}q+2)={a}_{1}+1+{a}_{1}{q}^{2}+2}\end{array}\right.$,
解之得$\left\{\begin{array}{l}{{a}_{1}=1}\\{q=2}\end{array}\right.$,或$\left\{\begin{array}{l}{{a}_{1}=4}\\{q=\frac{1}{2}}\end{array}\right.$(数列{an}为递增等比数列,舍去),
∴数列{an}的通项公式:an=2n-1;
(2)由bn=an+2n得,bn=2n-1+2n,
∴Tn=b1+b2+…+bn=(20+2×1)+(21+2×2)+(22+2×3)+…+(2n-1+2n)
=(20+21+22+…+2n-1)+2(1+2+3+…+n)
=$\frac{{2}^{0}(1-{2}^{n})}{1-2}$+2×$\frac{n(n+1)}{2}$
=2n+n2+n-1.
点评 本题考查了等差数列与等比数列的通项公式及其前n项和公式,考查了推理能力与计算能力,属于中档题.
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