第2个回答 2011-09-21
(1)因为an=5Sn-3,又S1=a1,所以a1=3/4,
Sn=an/5-3/5, S(n+1)=a(n+1)/5-3/5
a(n+1)=S(n+1)-Sn=a(n+1)/5-an/5
4a(n+1)/5=-an/5
a(n+1)/an=-1/4
所以数列(an)是等比数列,首项为3/4,公比为-1/4,通项为an=(3/4)(-1/4)^(n-1)
(2)令Bn=a(2n-1),则Bn是以B1=3/4为首项,q=(-1/4)^2=1/16的等比数列
a1+a3+。。。+a2n-1=B1+B2+……+Bn
=[(3/4)(1-1/16^n)]/(1-1/16)
=(12/15)(1-1/16^n)