Question

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！

Answer 1

(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) )

Answer 2

(1)b(n+1)=[a(n+2)*a(n+1)]^0.5,bn=[an*a(n+1)]^0.5
两式一比得 q=[a(n+2)/an]^0.5
即 a(n+2)=anq^2
（2）求an的通项 若n为偶数
则 a(n+2)/an=q^2
an/a(n-2)=q^2
....................
a4/a2=q^2
累乘得 an/a2=q^(n-2) an=2q^(n-2)
若n为奇数
则 a(n+2)/an=q^2
an/a(n-2)=q^2
.....................
a3/a1=q^2
累乘得 an=q^(n-1)
因此 Cn=q^(2(n-1))+2*2*q^(2(n-1))=5*(q^2)^(n-1)
故Cn是 首项为 5,公比为 q^2的等比数列
（3）1/a1+1/a2+1/a3+1/a4+…+1/a(2n-1)+1/a2n
=(1/a1+1/a3+....1/a(2n-1))+(1/a2+1/a4+....1/a2n)
=(1+1/q^2+1/q^4+......1/q^(2n-2))+1/2*(1+1/q^2+1/q^4+.........1/q^(2n-2))
=3/2*((1/q^2)^n-1)/(1/q^2-1)