数列412,814,1618,32116…,的前n项和为()A.2n+2-2-n-1B.2n+2-2-n-3C.2n+2+2-n-1D.2n+2-2-n-1-1-数学

题目简介

数列412,814,1618,32116…,的前n项和为()A.2n+2-2-n-1B.2n+2-2-n-3C.2n+2+2-n-1D.2n+2-2-n-1-1-数学

题目详情

数列4
1
2
,8
1
4
,16
1
8
,32
1
16
…,的前n项和为(  )
A.2n+2-2-n-1B.2n+2-2-n-3
C.2n+2+2-n-1D.2n+2-2-n-1-1
题型:单选题难度:偏易来源:不详

答案

解;根据题意设该数列为{an},前n项之和为Sn,
则an=2n+1+class="stub"1
2n

∴Sn=(4+8+16+…+2n+1)+(class="stub"1
21
+class="stub"1
22
+ class="stub"1
23
+…+class="stub"1
2n

=
4(1-2n)
1-2
+
class="stub"1
2
(1-class="stub"1
2n
)
1-class="stub"1
2
=2n+2-2-n-3.
故选B.

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