数列{an}的前n项和记为Sn,Sn=2an-2,(1)求{an}的通项公式
2个回答
s(n+1)=2a(n+1)-2
an=s(n+1)-sn
=2(an+1)-2-(2an-2)
2a(n+1)=3an
q=a(n+1)/an=3/2
a2=a1q=3a1/2
s2=a1+3a1/2=5a1/2
s2=2*3a1/2-2=3a1-2
5a1/2=3a1-2
a1=4
a2=6
a3=9
等差数列{bn}的各项为正,其前3项和为6
s3=3b2=6
b2=2
a1+b1,a2+b2,a3+b4成等比数列
即4+2-d,6+2,9+2+2d成等比数列
即6-d,8,11+2d成等比数列
(6-d)(11+2d)=64
2d^2-d-2=0
d^2-d/2-1=0
d^2-d/2+1/16-11/16=0
(d-1/4)^2-17/16=0
(d-1/4-√17/4)(d-1/4+√17/4)=0
d=1/4+√17/4 或 d=1/4-√17/4
当d=1/4+√17/4时
a2=a1+d
2=a1+1/4+√17/4
a1=2-(1/4+√17/4)=7/4-√17/4
an=7/4-√17/4+(n-1)(1/4+√17/4)
=7/4-√17/4-1/4-√17/4+(1/4+√17/4)n
=3/2-√17/2+(1/4+√17/4)n
当d=1/4-√17/4时
a2=a1+d
2=a1+1/4-√17/4
a1=2-(1/4-√17/4)=7/4+√17/4
an=7/4+√17/4+(n-1)(1/4-√17/4)
=7/4+√17/4-1/4+√17/4+(1/4-√17/4)n
=3/2+√17/2+(1/4-√17/4)n
相关问题
