已知数列{an}和{bn}满足：a1=1,a2=2,an>0,bn=跟号an*a(n+1)(n€R)

且{bn}是q为公比的等比数列.(1)证明： a(n+2)=anq^2; (2)若Cn=a(2n-1)+2a2n,证明数列{Cn}是等比数列； (3)求和：1/a1+1/a2+1/a3+1/a4+…+1/a(2n-1)+1/a2n 求详解尤其第三问，我好像算来1/an公比1/q^2,但与其他不同；求详解！谢谢！

(1)
bn=√(an.a(n+1)) (1)
b(n+1) =√(a(n+2).a(n+1)) (2)
(2)/(1)
√[a(n+2)/an] = bn/b(n-1)
=q
a(n+2)=q^2an
(2)

cn =a(2n-1) +2a(2n)
cn/c(n-1) =[a(2n-1) +2a(2n)]/[a(2n-3) +2a(2n-2)]
=q^2[a(2n-3) +2a(2n-2)]/[a(2n-3) +2a(2n-2)]
=q^2
=> {cn}是等比数列
(3)
√(an.a(n+1)) =√2.q^(n-1)
an.a(n+1) =2.q^(2n-2) (1)

a(n-1).an =2.q^(2n-4) (2)
(1)/(2)
a(n+1)/a(n-1) = q^2
an/a(n-2) =q^2
if n is odd
an/a1 = q^(n-1)
an = q^(n-1)

if n is even
an/a2 = q^(n-2)
an = 2.q^(n-2)

1/a1+1/a2+1/a3+1/a4+…+1/a(2n-1)+1/a(2n)
=1 + 1/2 + 1/q^2 + 1/(2q^2)+...+ 1/q^(2n-2) + 1/[2q^(2n-2)]
=(3/2)( 1+ 1/q^2+1/q^4+...+1/q^(2n-2) )
=(3/2) ( 1- q^(-2n)) /(1- q^(-2) )

由bn=an-1与an-1=an[(an+1)-1]得 bn=[bn+1]*（bn+1) 所以bn/[bn+1]=(bn+1) 所以[bn+1]/bn=1/(bn+1) 即1/bn+1=(bn+1) 所以{1/bn}是以b1=a1-1=1为首项，以1为公差的等差数列 所以1/bn=1+[n-1]*1=n 所以bn=1/n 其中[]为小括号，（）为下标，望采纳