若数列{an}的通项公式为an=n2n,则前n项和为()A.Sn=1-12nB.Sn=2-12n-1-n2nC.Sn=n(1-12n)D.Sn=2-12n-1+n2n-数学

题目简介

若数列{an}的通项公式为an=n2n,则前n项和为()A.Sn=1-12nB.Sn=2-12n-1-n2nC.Sn=n(1-12n)D.Sn=2-12n-1+n2n-数学

题目详情

若数列{an}的通项公式为an=
n
2n
,则前n项和为(  )
A.Sn=1-
1
2n
B.Sn=2-
1
2n-1
-
n
2n
C.Sn=n(1-
1
2n
D.Sn=2-
1
2n-1
+
n
2n
题型:单选题难度:偏易来源:不详

答案

可用错位相减求或验证S1、S2.
法一(验证法):S1=a1=class="stub"1
21
=class="stub"1
2
,排除D.
S2=a1+a2=class="stub"1
2
 +class="stub"2
22
=1
,排除A,C.选B
法二(错位相减法):Sn=a1+a2+…+an=class="stub"1
2
+class="stub"2
22
+…+class="stub"n
2n
,①
class="stub"1
2
Sn=class="stub"1
22
+class="stub"2
23
+…class="stub"n
2n+1
,②
①-②得:class="stub"1
2
Sn
=class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+…class="stub"1
2n
-class="stub"n
2n+1

Sn=1+class="stub"1
2
+class="stub"1
22
+…class="stub"1
2n-1
-class="stub"n
2n
=
1-(class="stub"1
2
)
n
1-class="stub"1
2
-class="stub"n
2n
=2-class="stub"1
2n-1
-class="stub"n
2n
,故选B.

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