问题 填空题
已知数列log2(an-1)(n∈N*)为等差数列,且a1=3,a2=5,则
1
a2-a1
+
1
a3-a2
+…+
1
an+1-an
=______.
答案

设等差数列的公差为d,则d=log2(a2-1)-log2(a1-1)=1

∴log2(an-1)=log22+(n-1)×1=n

∴an=2n+1

则an+1-an=2n+1-2n=2n

1
a2-a1
+
1
a3-a2
+…+
1
an+1-an
=
1
2
+
1
22
+…+
1
2n
=
1
2
[1-(
1
2
)]
n
1-
1
2
=1-
1
2n

故答案为:1-

1
2n

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