已知数列{log2(an−1)}(n∈N+)为等差数列,且a1=3,a2=5,则1a2−a1+1a3−a2+…+1an+
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已知数列{log
令bn=log2(an-1),(n∈N+),依题意{bn}为等差数列,
∵a1=3,a2=5,
∴b1=log2(3-1)=1,b2=log2(5-1)=2,
∵{bn}为等差数列,设其公差为d,则d=1,
∴bn=n,
∴an=2n+1,
∴
1
an+1−an=
1
(2n+1+1)−(2n+1)=
1
2n,
显然{
1
2n}是首项为
1
2,公比为
1
2的等比数列,
∴
1
a2−a1+
1
a3−a2+
1
a4−a3+…+
1
an+1−an=
1
2+
1
22+
1
23+…+
1
2n
=
1
2(1−(
1
2)n)
1−
1
2=1-(
1
2)n.
故选C.
∵a1=3,a2=5,
∴b1=log2(3-1)=1,b2=log2(5-1)=2,
∵{bn}为等差数列,设其公差为d,则d=1,
∴bn=n,
∴an=2n+1,
∴
1
an+1−an=
1
(2n+1+1)−(2n+1)=
1
2n,
显然{
1
2n}是首项为
1
2,公比为
1
2的等比数列,
∴
1
a2−a1+
1
a3−a2+
1
a4−a3+…+
1
an+1−an=
1
2+
1
22+
1
23+…+
1
2n
=
1
2(1−(
1
2)n)
1−
1
2=1-(
1
2)n.
故选C.
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