Question

已知数列{a_n}的前n项和S_n满足S_n=2ⁿ+1，则当n≥2时，

1

a1

+

1

a2

+…+

1

an

=______．

Answer 1

由S_n=2ⁿ+1 得，当n=1时，a₁=S₁=3，当n≥2时a_n=Sn-Sn-1=2 ^n-1∴an=

3   n=1 2n-1  n＞1

1

a1

+

1

a2

+…+

1

an

=

1

3

+

1

2

+

1

22

+…+

1

2n-1

=

1

3

+

1 2 [1-( 1 2 )n-1]

1- 1 2

=

4

3

-(

1

2

)n-1
故答案为：

4

3

-(

1

2

)n-1