1.等比数列｛an}中,a1+a3=10,a4+a6=5/4,则数列｛an}的通项公式为A.an=2^4-nB.an=2^n-4C.an=2^n-3D.an=2^3-n2设sn是等比数列｛an}的前n项和,s4/S6=1/3,则S6/S12=3.数列｛an}{bn}满足anbn=1,an=n^2+3n+2,则{bn}的前10项之和为

Question

1.等比数列｛an}中,a1+a3=10,a4+a6=5/4,则数列｛an}的通项公式为
A.an=2^4-n B.an=2^n-4 C.an=2^n-3 D.an=2^3-n
2设sn是等比数列｛an}的前n项和,s4/S6=1/3,则S6/S12=
3.数列｛an}{bn}满足anbn=1,an=n^2+3n+2,则{bn}的前10项之和为

Answer 1

A.an=2^(4-n )
a4+a6
=a1q^3+a2q^3
=q^3(a1+a3)
5/4=q^3*10
q^3=1/8
q=1/2
a1+a3
=a1+a1q^2
=a1(1+q^2)
10=a1*(1+1/2^2)
10=a1*5/4
a1=8
an=a1q^(n-1)
=8*(1/2)^(n-1)
=(1/2)^(n-4)
=2^(4-n)
因为Sn是等比数列{an}的前n项和
所以S3,S6-S3,S9-S6,S12-S9也是等比数列
又S3/S6=1/3
所以S6=3S3
S6-S3=2S3
所以公比是(S6-S3)/S3=2S3/S3=2
所以S9-S6=2(S6-S3)=2*2S3=4S3
故S9=S6+4S3=3S3+4S3=7S3
S12-S9=2(S9-S6)=2*4S3=8S3
故S12=S9+8S3=7S3+8S3=15S3
所以S6/S12=2S3/15S3=2/15
an=n^2+3n+2
an=(n+1)(n+2)
anbn=1
bn=1/an=1/[(n+1)(n+2)]
=[(n+2)-(n+1)]/[(n+1)(n+2)]
=(n+2)/[(n+1)(n+2)]-(n+1)/[(n+1)(n+2)]
=1/(n+1)-1/(n+2)
所以S10=(1/2-1/3)+(1/3-1/4)+……+(1/11-1/12)
=1/2-1/12
=6/12-1/12
=5/12