设数列{an}满足a1=1,a2+a4=6,且对任意n∈N*,函数f(x)=(an-an+1+an+2)x+an+1•cosx-an+2sinx满足f′(π2)=0若cn=an+12an,则数列{cn

题目简介

设数列{an}满足a1=1,a2+a4=6,且对任意n∈N*,函数f(x)=(an-an+1+an+2)x+an+1•cosx-an+2sinx满足f′(π2)=0若cn=an+12an,则数列{cn

题目详情

设数列{an}满足a1=1,a2+a4=6,且对任意n∈N*,函数f(x)=(an-an+1+an+2)x+an+1•cosx-an+2sinx满足f′(
π
2
)=0
cn=an+
1
2an
,则数列{cn}的前n项和Sn为(  )
A.
n2+n
2
-
1
2n
B.
n2+n+4
2
-
1
2n-1
C.
n2+n+2
2
-
1
2n
D.
n2+n+4
2
-
1
2n
题型:单选题难度:中档来源:不详

答案

∵f(x)=(an-an+1+an+2)x+an+1•cosx-an+2sinx,
∴f′(x)|x=class="stub"π
2
=an-an+1+an+2-an+1•sinx|x=class="stub"π
2
-an+2cosx|x=class="stub"π
2

=an-2an+1+an+2,
∵f′(class="stub"π
2
)=0,
∴an-2an+1+an+2=0,即2an+1=an+an+2,
∴数列{an}是等差数列,设其公差为d,
∵a2+a4=6,
∴2a1+4d=6,a1=1,
∴d=1,
∴an=1+(n-1)×1=n,
∴cn=an+class="stub"1
2an
=n+class="stub"1
2n

∴Sn=c1+c2+…+cn
=(1+2+…+n)+(class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n

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

=
n2+n+2
2
-class="stub"1
2n

故选:C.

更多内容推荐