题目内容
4.已知各项均为正数的数列{an}的前n项和Sn满足S1>1,且6Sn=(an+1)(an+2)(n为正整数).设数列{bn}满足bn=$\left\{\begin{array}{l}{{a}_{n},n为偶数}\\{{2}^{{a}_{n}},n为奇数}\end{array}\right.$,求Tn=b1+b2+…+bn.分析 由6Sn=(an+1)(an+2)(n为正整数),n=1时,6a1=a12+3a1+2,且a1>1,解得a1.n≥2时,6Sn=an2+3an+2,6Sn-1=an-12+3an-1+2,相减可得:an-an-1=3,可得an=3n-1.由bn=$\left\{\begin{array}{l}{{a}_{n},n为偶数}\\{{2}^{{a}_{n}},n为奇数}\end{array}\right.$,通过分类讨论,利用等差数列与等比数列的求和公式即可得出.
解答 解:n=1时,6a1=a12+3a1+2,且a1>1,解得a1=2.
n≥2时,6Sn=an2+3an+2,6Sn-1=an-12+3an-1+2,
两式相减得:6an=an2-an-12+3an-3an-1
即(an+an-1)(an-an-1-3)=0,
∵an+an-1>0,
∴an-an-1=3,
∴{an}为等差数列,an=3n-1.
由bn=$\left\{\begin{array}{l}{{a}_{n},n为偶数}\\{{2}^{{a}_{n}},n为奇数}\end{array}\right.$,
当n为偶数时,Tn=(b1+b3+…+bn-1)+(b2+b4+…+bn)
=$\frac{4({8}^{\frac{n}{2}}-1)}{8-1}$+$\frac{\frac{n}{2}(5+3n-1)}{2}$=$\frac{4}{7}({8}^{\frac{n}{2}}-1)$+$\frac{n(3n+4)}{3}$,
当n为奇数时,Tn=(b1+b3+…+bn)+(b2+b4+…+bn-1)
=$\frac{4({8}^{\frac{n+1}{2}}-1)}{8-1}$+$\frac{\frac{n-1}{2}(5+3n-4)}{2}$
=$\frac{4}{7}({8}^{\frac{n+1}{2}}-1)$+$\frac{(n-1)(3n+1)}{4}$.
∴Tn=$\left\{\begin{array}{l}{\frac{4}{7}({8}^{\frac{n}{2}}-1)+\frac{n(3n+4)}{3},n为偶数}\\{\frac{4}{7}({8}^{\frac{n+1}{2}}-1)+\frac{(n-1)(3n+1)}{4},n为奇数}\end{array}\right.$.
点评 本题考查了等比数列的通项公式与求和公式、递推关系,考查了分类讨论方法、推理能力与计算能力,属于中档题.
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