在等差数列{an}中,a2=4,a6=12,,那么数列{an2n+1}的前n项和等于()A.2-n+22nB.1+n+12nC.1+n2nD.n(n-1)2n+1-数学

题目简介

在等差数列{an}中,a2=4,a6=12,,那么数列{an2n+1}的前n项和等于()A.2-n+22nB.1+n+12nC.1+n2nD.n(n-1)2n+1-数学

题目详情

在等差数列{an}中,a2=4,a6=12,,那么数列{
an
2n+1
}的前n项和等于(  )
A.2-
n+2
2n
B.1+
n+1
2n
C.1+
n
2n
D.
n(n-1)
2n+1
题型:单选题难度:中档来源:不详

答案

∵等差数列{an}中,a2=4,a6=12;
∴公差d=
a6a2
6-2
=class="stub"12-4
6-2
=2

∴an=a2+(n-2)×2=2n;
an
2n+1
=class="stub"n
2n

an
2n+1
的前n项和,
Sn=1×class="stub"1
2
+2×(class="stub"1
2
)
2
+3×(class="stub"1
2
)
3
+…
+(n-1)×(class="stub"1
2
)
n-1
+n×(class="stub"1
2
)
n

class="stub"1
2
Sn
=(class="stub"1
2
)
2
+2×(class="stub"1
2
)
3
+3×(class="stub"1
2
)
4
…+(n-1)×(class="stub"1
2
)
n
+n×(class="stub"1
2
)
n+1

两式相减得class="stub"1
2
Sn=class="stub"1
2
+(class="stub"1
2
)
2
+(class="stub"1
2
)
3
+…+(class="stub"1
2
)
n
-n(class="stub"1
2
)
n+1

=
class="stub"1
2
-(class="stub"1
2
)
n+1
1-class="stub"1
2
- n(class="stub"1
2
)
n+1

Sn=1+class="stub"n+1
2n

故选B

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