0,bn=根号anan+1,且{bn}是以q为公比的等比数列.1.证明：an+2=anq22.若cn

0,bn=根号anan+1,且{bn}是以q为公比的等比数列.1.证明：an+2=anq22.若cn

b1=√a1a2=√2b2=b1q=√a2a3,a3=b1^2q^2/a2=q^2bn=b1q^(n-1)=√anan+1bn+2=b1q^(n+1)=√an+1an+2anan+1=2q^(n-1)an+2an+1=2q^(n+1)an/an+2=1/q^2an+2=an *q^21、得证2、cn=a(2n-1)+2a(2n)a(2n+2)=q^2a(2n)a(2n+1)=a(2n-1+2)=q^2a(2n-1)cn+1/cn=[a(2n-1+2)+2a(2n+2)]/[a(2n-1)+2a(2n)]=q^2*[a(2n-1)+2a(2n)]/[a(2n-1)+2a(2n)]=q^2∴ {cn}是等比数列,公比q^23、an+2=anq^21/a(2n)=1/a(2n-2+2)=1/q^2a(2n-2)=1/q^4a(2n-4)=1/q^6a(2n-6)=1/[q^2(n-1)a(2n-2n+2]同理,1/a(2n-1)=1/q^2a(2n-3)=1/q^2(n-1)a1S=1/a1+1/a2+...+1/a(2n-1)+1/a(2n)是两个等比数列之和,公比都是q^2,第一项分别是b1=1/a1=1,c1=1/a2=1/2都是n项据求和公式：S=(1-q^2n)/(1-q^2)+(1/2)(1-q^2n)/(1-q^2)=(3/2)(1-q^n)(1+q^n)/(1-q)(1+q)q≠±1 q^2=1则,a3=a1=a5=...=a(2n-1)=1a2=a4=a6=...=a(2n)=1/2S=n/2+n=3n/2这是我在静心思考后得出的结论,如果不能请追问,我会尽全力帮您解决的~如果您有所不满愿意,请谅解~

哦，回答的不错