问题
选择题
已知函数f(x)=x2-bx的图象在点A(1,f(1))处的切线l与直线x+3y-1=0垂直,若数列{
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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l与直线x+3y-1=0垂直,∴-b+2=3
∴b=-1,f(x)=x2+x
∴f(n)=n2+n=n(n+1)
∴
=1 f(n)
=1 n(n+1)
-1 n 1 n+1
∴S2012=
+1 f(1)
+…+1 f(2)
=1-1 f(2012)
+1 2
-1 2
+…+1 3
-1 2012
=1-1 2013
=1 2013 2012 2013
故选A.