数列1+12,2+14,3+18,4+116,…的前n项和为(  ) A.2−12n−n2n+1 B.2−12n−1−n2n C.n2(n+1)+1−12n D.n(n+1)2+1−12n−1

问题描述:

数列1+

1
2
2+
1
4
3+
1
8
4+
1
16
,…的前n项和为(  )
A. 2−
1
2n
n
2n+1

B. 2−
1
2n−1
n
2n

C.
n
2
(n+1)+1−
1
2n

D.
n(n+1)
2
+1−
1
2n−1

1+

1
2
+2+
1
4
+3+
1
8
+4+
1
16
+…+n+
1
2n

=(1+2+3+…+n)+(
1
2
+
1
4
+…+
1
2n

=
(1+n)n
2
+
1
2
(1−(
1
2
)n)
1−
1
2

=
n
2
(n+1)+1−
1
2n

故选:C.