数列1+12,2+14,3+18,4+116,…的前n项的和为()A.12n+n2+n2B.-12n+n2+n2+1C.-12n+n2+n2D.-12n+1+n2-n2-数学

题目简介

数列1+12,2+14,3+18,4+116,…的前n项的和为()A.12n+n2+n2B.-12n+n2+n2+1C.-12n+n2+n2D.-12n+1+n2-n2-数学

题目详情

数列1+
1
2
,2+
1
4
,3+
1
8
,4+
1
16
,…的前n项的和为(  )
A.
1
2n
+
n2+n
2
B.-
1
2n
+
n2+n
2
+1
C.-
1
2n
+
n2+n
2
D.-
1
2n+1
+
n2-n
2
题型:单选题难度:中档来源:不详

答案

数列1+class="stub"1
2
,2+class="stub"1
4
,3+class="stub"1
8
,4+class="stub"1
16
,…的通项公式为n+(class="stub"1
2
)
n

∴则该数列的前n项的和为1+2+3+…+n+class="stub"1
2
+class="stub"1
4
+…+(class="stub"1
2
)
n
=-class="stub"1
2n
+
n2+n
2
+1

故选B.

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