数列1,2+12,3+12+14,…,n+12+14+…+12n-1的前n项和为()A.n+1-(12)n-1B.12n2+32n+12n-1-3C.12n2+32n+12n-1-2D.n+12n-1

题目简介

数列1,2+12,3+12+14,…,n+12+14+…+12n-1的前n项和为()A.n+1-(12)n-1B.12n2+32n+12n-1-3C.12n2+32n+12n-1-2D.n+12n-1

题目详情

数列1,2+
1
2
,3+
1
2
+
1
4
,…,n+
1
2
+
1
4
+…+
1
2n-1
的前n项和为(  )
A.n+1-(
1
2
)
n-1
B.
1
2
n2+
3
2
n+
1
2n-1
-3
C.
1
2
n2+
3
2
n+
1
2n-1
-2
D.n+
1
2n-1
-1
题型:单选题难度:偏易来源:不详

答案

∵此数列的通项an=n+class="stub"1
2
+class="stub"1
4
+…+class="stub"1
2n-1
=n+
class="stub"1
2
[1-(class="stub"1
2
)n-1]
1-class="stub"1
2
=n+1-class="stub"1
2n-1

∴此数列的前n项和Sn=2+3+…+(n+1)-1-class="stub"1
2
-class="stub"1
4
-…-class="stub"1
2n-1
=
n(n+3)
2
-
1×[1-(class="stub"1
2
)n]
1-class="stub"1
2
=class="stub"1
2
n2+class="stub"3
2
n
-2+class="stub"1
2n-1

故选C.

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