Question

数列S

1

2

+

3

22

+

5

23

…+

2n-1

2n

的前n项和为 ______．

Answer 1

S_n=

1

2

+

3

22

+

5

23

…+

2n-1

2n

1

2

S_n=

1

22

+

3

23

+

5

24

+…+

2n-2

2n

+

2n-1

2n+1

两式相减得

1

2

S_n=

1

2

+

2

22

+

2

23

+…+

2

2n

-

2n-1

2n+1

=

1

2

+

1 4 (1- 1 2n-1 )

1- 1 2

-

2n-1

2n+1

=

3

2

-

2n-1

2n+1

∴S_n=3-

2n+3

2n

故答案为：3-

2n+3

2n