函数f(x)=x2+bx在点A(1,f(1))处的切线方程为3x-y-1=0,设数列{1f(n)}的前n项和为Sn,则S2012为______.-数学

题目简介

函数f(x)=x2+bx在点A(1,f(1))处的切线方程为3x-y-1=0,设数列{1f(n)}的前n项和为Sn,则S2012为______.-数学

题目详情

函数f(x)=x2+bx在点A(1,f(1))处的切线方程为3x-y-1=0,设数列{
1
f(n)
}
的前n项和为Sn,则S2012为______.
题型:填空题难度:偏易来源:不详

答案

∵f(x)=x2+bx
∴f′(x)=2x+b
∴y=f(x)的图象在点A(1,f(1))处的切线斜率k=f′(1)=2+b
∵切线与直线3x-y+2=0平行
∴b+2=3
∴b=1,f(x)=x2+x
∴f(n)=n2+n=n(n+1)
class="stub"1
f(n)
=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

∴S2012=class="stub"1
f(1)
+class="stub"1
f(2)
+…+class="stub"1
f(2012)
=1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+class="stub"1
3
+…+class="stub"1
2012
-class="stub"1
2013
=1-class="stub"1
2013
=class="stub"2012
2013

故答案为 class="stub"2012
2013

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