Question

数列1+

1

2

 ， 2+

1

4

 ， 3+

1

8

 ， … ， n+

1

2n

 ， …的前n项和是______．

Answer 1

Sn=1+

1

2

+2+

1

4

+…+n+

1

2n

=(1+2+…+n)+(

1

2

+

1

4

+…+

1

2n

)
=

n(n+1)

2

+

1 2 [1-( 1 2 )n]

1- 1 2

=

n(1+n)

2

+1-(

1

2

)n
故答案为：

n(n+1)

2

+1-(

1

2

)n