已知数列满足:a1=1,a(n+1)=an+1,n为奇数;2an,n为偶数,设bn=a2n-1,(Ⅰ)求b2,b3,并证明：b(n+1)=2bn+2;(Ⅱ)①证明：数列｛bn+2}为等比数列；②若a2k,a(2k+1),9+a(2k+2)成等比数列,求正整数k的值.

(I)
a(n+1)=an +1 ,n is odd
=2an ,n is even
bn=a(2n) -1
a1=1
a2 =a1+1 =2
if n is odd,
a(n+1) = an +1
= 2a(n-1) +1
a(n+1) +1 = 2[ (a(n-1) +1 ]
a(n+1) +1 = 2^[( n-1)/2 ].(a2 +1 )
= 3.2^[( n-1)/2 ]
a(n+1) = -1+3.2^[( n-1)/2 ]
n is odd => n= 2m-1
a(2m) =-1+3.2^(m-1)
bn = a(2n) -1
=-2+3.2^(n-1)
b2 = -2+3.2 = 4
b3= -2+3.4= 10
b(n+1) =-2+3.2^n
=2(-2+3.2^(n-1)) +2
=2bn +2
(II)
(1)
bn+2 =3.2^(n-1)
{bn+2}是等比数列,q=3
(2)
if n is even,
a(n+1)=2an
= 2(a(n-1)+1)
a(n+1)+2 =2(a(n-1)+2)
= 2^(n/2).(a1+2)
=3.2^(n/2)
a(n+1) =-2+3.2^(n/2)
n is even ,n=2k
a(2k+1) =-2+3.2^k
a(2k),a(2k+1),9+a(2k+2)成等比数列
a(2k).[9+a(2k+2)]= [a(2k+1)]^2
[-1+3.2^(k-1) ].(8+3.2^k ) =(-2+3.2^k)^2
-8-3.2^k+12.2^k +9.2^(2k-1) = 4 -12.2^k + 9.2^(2k)
(9/2).2^(2k) -21.2^k +12=0
9.2^(2k) -42.2^k +24=0
(2^k -4)(9.2^k-6)=0
2^k=4
k=2