B.16
C.24
D.48
【解析】由程序框图可列表如下:
n 6 12 24 S 3 3-3
因为3-3≈3.106>3.10,所以输出n的值为24,故选C.
6.已知数列的前n项和为Sn,通项公式an=log2(n∈N*),则满足不等式Sn<-6的n的最小值是(D)
A.62 B.63 C.126 D.127
【解析】因为Sn=log2=log2<-6,所以<2-6,n>126,故应选D.
7.设A、B、C为圆O上三点,且AB=3,AC=5,则\s\up6(→(→)·\s\up6(→(→)=(D)
A.-8 B.-1 C.1 D.8
【解析】取BC的中点D,连接AD,OD,因为O为三角形ABC外接圆的圆心,则\s\up6(→(→)=(\s\up6(→(→)+\s\up6(→(→)),\s\up6(→(→)·\s\up6(→(→)=0.所以\s\up6(→(→)·\s\up6(→(→)=(\s\up6(→(→)+\s\up6(→(→))·\s\up6(→(→)=\s\up6(→(→)·\s\up6(→(→)=(\s\up6(→(→)+\s\up6(→(→))·(\s\up6(→(→)-\s\up6(→(→))=(|\s\up6(→(→)|2-|\s\up6(→(→)|2)=8,选D.
8.已知定义在R上的奇函数f(x)满足f(x)=f(x+2),数列的前n项和为Sn,且Sn=2an+2,则f(an)=(A)
A.0 B.0或1 C.-1或0 D.1或-1
【解析】∵f(x)=f(x+2),所以f(x)函数周期为2,∵数列满足Sn=2an+2,∴a1=-2,Sn-1=2an-1+2,∴an=2an-2an-1,即an=2an-1,∴{an}以-2为首项,2为公比的等比数列,∴an=-2n,∴f(an)=f(-2n)=f=0,故选A.