D3等比数列及等比数列前n项和【数学理卷·届辽宁省沈阳二中高三上学期期中考试(11)】19.(本题满分12分)设数列是等差数列,数列的前项和满足且(Ⅰ)求数列和的通项公式:(Ⅱ)设,设为的前n项和,求
【知识点】等差数列等比数列数列求和D2D3D4【答案解析】(1),
(2)(1) 数列{bn}的前n项和Sn满足Sn=(bn-1),∴b1=S1=(b1-1),解得b1=3.当n≥2时,bn=Sn-Sn-1=(bn-1)-(bn-1-1),化为bn=3bn-1.∴数列{bn}为等比数列,∴bn=3×3n-1=3n. a2=b1=3,a5=b2=9.设等差数列{an}的公差为d.∴,解得d=2,a1=1.∴an=2n-1.综上可得:an=2n-1,bn=3n.(2)cn=an•bn=(2n-1)•3n.∴Tn=3+3×32+5×33+…+(2n-3)•3n-1+(2n-1)•3n,3Tn=32+3×33+…+(2n-3)•3n+(2n-1)•3n+1.∴-2Tn=3+2×32+2×33+…+2×3n-(2n-1)•3n+1=-(2n-1)•3n+1-3=(2-2n)•3n+1-6.∴Tn=3+(n-1)3n+1.【思路点拨】(1)利用等差数列与等比数列的通项公式即可得出;(2)利用“错位相减法”和等比数列的前n项和公式即可得出.【数学理卷·届辽宁省沈阳二中高三上学期期中考试(11)】8.已知等比数列满足>0,=1,2,…,且,则当≥1时,=()A.n(2n-1)B.(n+1)2C.n2D.(n-1)2【知识点】等比数列及等比数列前n项和D3【答案解析】C由等比数列的性质可得an2=a5•a2n-5=22n,=(2n)2, an>0,∴an=2n,故数列首项a1=2,公比q=2,故log2a1+log2a3+…+log2a2n-1=log2a1•a3•…•a2n-1=log2(a