已知数列{log2(an-1)}(n∈N*)为等差数列,且a1=3,a3=9. (Ⅰ)求数列{an}的通项公式; (Ⅱ)证明1/a2−a1+1/a3−a2+…+1/an+1−an<1.
问题描述:
已知数列{log2(an-1)}(n∈N*)为等差数列,且a1=3,a3=9.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)证明
+1
a2−a1
+…+1
a3−a2
<1. 1
an+1−an
答
(I)设等差数列{log2(an-1)}的公差为d.
由a1=3,a3=9得2(log22+d)=log22+log28,即d=1.
所以log2(an-1)=1+(n-1)×1=n,即an=2n+1.
(II)证明:因为
=1
an+1−an
=1
2n+1−2n
,1 2n
所以
+1
a2−a1
+…+1
a3−a2
=1
an+1−an
+1 21
+1 22
+…+1 23
=1 2n
=1-
−1 2
×1 2n
1 2 1−
1 2
<1,1 2n
即得证.