大题冲关集训(三)1
(2014哈尔滨一模)数列{an}满足an+1-an=2,a1=2,等比数列{bn}满足b1=a1,b4=a8
(1)求数列{an},{bn}的通项公式;(2)设cn=anbn,求数列{cn}的前n项和Tn
解:(1)an+1-an=2,a1=2,所以数列{an}为等差数列,则an=2+(n-1)×2=2n,b1=a1=2,b4=a8=16,所以q3==8,q=2,则bn=2n
(2)cn=anbn=n·2n+1,则Tn=1×22+2×23+3×24+…+n·2n+1,2Tn=1×23+2×24+3×25+…+n·2n+2,两式相减得-Tn=1×22+23+24+…+2n+1-n·2n+2,整理得Tn=(n-1)2n+2+4
(2013高考福建卷)已知等差数列{an}的公差d=1,前n项和为Sn
(1)若1,a1,a3成等比数列,求a1;(2)若S5>a1a9,求a1的取值范围
解:(1)因为数列{an}的公差d=1,且1,a1,a3成等比数列,所以=1×(a1+2),即-a1-2=0,解得a1=-1或a1=2
(2)因为数列{an}的公差d=1,且S5>a1a9,所以5a1+10>+8a1,即+3a1-10
