已知函数f(x)=x2+2bx的图象在点A(0,f(0))处的切线L与直线x-y+3=0平行,若数列{1f(n)}的前n项和为Sn,则S2013的值为()A.20102011B.20112012C.2

题目简介

已知函数f(x)=x2+2bx的图象在点A(0,f(0))处的切线L与直线x-y+3=0平行,若数列{1f(n)}的前n项和为Sn,则S2013的值为()A.20102011B.20112012C.2

题目详情

已知函数f(x)=x2+2bx的图象在点A(0,f(0))处的切线L与直线x-y+3=0平行,若数列{
1
f(n)
}的前n项和为Sn,则S2013的值为(  )
A.
2010
2011
B.
2011
2012
C.
2012
2013
D.
2013
2014
题型:单选题难度:中档来源:不详

答案

由题意得,f′(x)=2x+2b,
∵在点A(0,f(0))处的切线L与直线x-y+3=0平行,
∴f′(0)=2b=1,得b=class="stub"1
2

∴f(x)=x2+x,
class="stub"1
f(n)
=class="stub"1
n2+n
=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

∴S2013=(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
2013
-class="stub"1
2014
)]
=1-class="stub"1
2014
=class="stub"2013
2014

故选D.

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