Question

已知数列｛log2(an-1)}(n属于N＊）为等差数列，且a1=3,a3=9,求数列｛an}的通项公式．

已知数列｛log2(an-1)}(n属于N＊）为等差数列，且a1=3,a3=9,求数列｛an}的通项公式．
请详细说一下过程～谢谢！

Answer 1

d=log2(an-1)-log2(a(n-1)-1)
=log2((an-1)/(a(n-1)-1))
则p=(an-1)/(a(n-1)-1)为定值
n=2 p=(a2-1)/2
n=3 p=8/(a2-1)
a2=5(an大于1)
p=2,d=1
2=(an-1)/(a(n-1)-1)
an=2a(n-1)-1
an=2^n+1

Answer 2

log2(a1-1)=log2(2)=1
log2(a3-1)-log2(a1-1)=2d
d=[log2(8)-log2(2)]/2=1
log2(an-1)=1+(n-1)*1=n
2^n=an-1
an=2^n+1

Answer 3

d=log2(an-1)-log2(a(n-1)-1)
=log2((an-1)/(a(n-1)-1))
则p=(an-1)/(a(n-1)-1)为定值
n=2 p=(a2-1)/2
n=3 p=8/(a2-1)
a2=5(an大于1)
p=2,d=1
2=(an-1)/(a(n-1)-1)
an=2a(n-1)-1
an=2^n+1

Answer 4

设bn＝n
㏒2﹙an－1﹚＝n
an－1＝2^n
an＝2^n＋1

Answer 5

log2\(A1-1)=log2\2=1,log2\(A3-1)=1+2d=3,d=1,
log2\(An-1)=1+(n-1)*1=n,
An-1=2^n,
An=2^n+1.