问题 选择题
已知数列{log2(an-1)}(n∈N*)为等差数列,且a1=3,a2=5,则
lim
n→∞
1
a2-a1
+
1
a3-a2
+…+
1
an+1-an
)=(  )
A.2B.
3
2
C.1D.
1
2
答案

数列{log2(an-1)}(n∈N*)为等差数列,

设其公差为d,则log2(an-1)-log2(an-1-1)=d,

an-1
an-1-1
=2d,又由a1=3,a2=5,

则d=1,即

an-1
an-1-1
=2,

{an-1}是以a1-1=2为首项,公比为2的等比数列,

进而可得,an-1=2n,则an=2n+1,

故an-an-1=2n-2n-1=2n-1

lim
n→∞
1
a2-a1
+
1
a3-a2
+…+
1
an+1-an
)=
lim
n→∞
1
2
+
1
4
+…+
1
2n-1
)=1,

故选C.

计算题
单项选择题