Question

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

(2)证明:1/(a2-a1)+1/(a3-a2)+…+1/(an+1-an)<1

Answer 1

1.
log2 (an-1)=log2 (a1-1)+(n-1)d=log2 2+(n-1)d=1+(n-1)d
1og2 (a3-1)=1+2d=log2 8=3
d=1
log2 (an-1)=1+(n-1)=n
an-1=2^n
an=1+2^n

2.
1/(a2-a1)+1/(a3-a2)+…+1/[a(n+1)-an]
=1/(2^2-2^1)+1/(2^3-2^2)+…+1/[2^(n+1)-2^n]
=1/2^1+1/2^2+…+1/2^n
=(1/2)(1-1/2^n)/(1-1/2)
=1-1/2^n
＜1

Answer 2

(1)设bn=log<2>(an-1),依题意
b1=1,b3=3,
{bn}为等差数列，d=1,bn=n,
∴an-1=2^n,
an=2^n+1.
(2)a<n+1>-an=2^(n+1)-2^n=2^n,
∴1/(a2-a1)+1/(a3-a2)+…+1/(an+1-an)
=1/2+1/2^2+……+1/2^n
=1-1/2^n<1.

Answer 3

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
2
1/(a2-a1)+1/(a3-a2)+…+1/[a(n+1)-an]
=1/(2^2-2^1)+1/(2^3-2^2)+…+1/[2^(n+1)-2^n]
=1/2^1+1/2^2+…+1/2^n
=(1/2)(1-1/2^n)/(1-1/2)
=1-1/2^n小于1