1.
a(n+1)=pan+2ⁿ
a(n+2)=pa(n+1)+2^(n+1)=p(pan+2ⁿ)+2^(n+1)=p²an+(p+2)×2ⁿ
数列是等比数列,则
a(n+1)²=an×a(n+2)
(pan+2ⁿ)²=an×[p²an+(p+2)×2ⁿ]
整理,得
(2-p)an=2ⁿ
n=1 a1=2代入
(2-p)×2=2
2-p=1
p=1
an=2ⁿ
2.
设公比为q
q=a(n+1)/an=2^(n+1)/2ⁿ=2
新数列依次是原数列的第2、3、5、6、……项.奇数项是以a2为首项,q³为公比的等比数列；偶数项是以a3为首项,q³为公比的等比数列.
n为奇数时,
bn=a2(q³)^[(n-1)/2]=a1q(q³)^[(n-1)/2]
=a1q^[(3n-1)/2]
=2×2^[(3n-1)/2]
=2^[(3n+1)/2]
n为偶数时,
bn=a3(q³)^[(n-2)/2]=a1q²(q³)^[(n-2)/2]
=a1q^[(3n-2)/2]
=2×2^[(3n-2)/2]
=2^(3n/2)
化为统一形式：
bn=2^[(3n/2)+(1/4)-(-1)ⁿ×(1/4)]