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+1?an=1-12n1-12n

2025-05-16 17:11:46
推荐回答(1个)
回答1:

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