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+
,2+1 2
,3+1 4
,4+1 8
,…的前n项和为( )1 16
A. 2−
−1 2n
n 2n+1
B. 2−
−1 2n−1
n 2n
C.
(n+1)+1−n 2
1 2n
D.
+1−n(n+1) 2
1 2n−1
答
1+
+2+1 2
+3+1 4
+4+1 8
+…+n+1 16
1 2n
=(1+2+3+…+n)+(
+1 2
+…+1 4
)1 2n
=
+(1+n)n 2
(1−(1 2
)n)1 2 1−
1 2
=
(n+1)+1−n 2
.1 2n
故选:C.