问题
选择题
数列1,2+
|
答案
∵此数列的通项an=n+
+1 2
+…+1 4
=n+1 2n-1
=n+1-
[1-(1 2
)n-1]1 2 1- 1 2
.1 2n-1
∴此数列的前n项和Sn=2+3+…+(n+1)-1-
-1 2
-…-1 4
=1 2n-1
-n(n+3) 2
=1×[1-(
)n]1 2 1- 1 2
n2+1 2
n-2+3 2
.1 2n-1
故选C.