1=√a1a2=√2
b2=b1q=√a2a3,a3=b1^2q^2/a2=q^2
bn=b1q^(n-1)=√anan+1
bn+2=b1q^(n+1)=√an+1an+2
anan+1=2q^(n-1)
an+2an+1=2q^(n+1)
an/an+2=1/q^2
an+2=an *q^2
1、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^2
3、
an+2=anq^2
1/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)a1
S=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)=1
a2=a4=a6=...=a(2n)=1/2
S=n/2+n=3n/2