已知函数f(x)=ax^2+bx+c的图像在点x=1处的切线L为直线3x-y-1=0,Tn=f(n)为等差数列{an}的前n项和,若数列{1/f(n)}的前n项和为Sn,则S2013的值为什么?

Question

已知函数f(x)=ax^2+bx+c的图像在点x=1处的切线L为直线3x-y-1=0,Tn=f(n)为等差数列{an}的前n项和,若数列{1/f(n)}的前n项和为Sn,则S2013的值为什么?

Answer 1

f'(x)=2ax + b,
f(x)在x=1处的切线为 3x-y-1=0,y = 3x-1.
切点(1,2)在f(x)图像上.因此,
f(1)=a+b+c=2.
切点(1,2)处切线的斜率为3,f'(1)=2a+b=3.b = 3-2a.
c = 2-a-b = 2-a-(3-2a)=a-1.
f(x)=ax^2 + (3-2a)x + (a-1).
f(n) = an^2 + (3-2a)n + (a-1),
a(n) = g + (n-1)d,f(n) = ng + n(n+1)d/2.
0 = f(0) = a-1,a=1.
f(n) = n^2 + n,
a(1)=f(1)=2,
a(n+1)=f(n+1)-f(n) = (n+1)^2 + (n+1) - n^2 - n = (2n+1) + 1 = 2(n+1),
a(n) = 2n.
h(n) = 1/f(n) = 1/[n(n+1)] = 1/n - 1/(n+1),
s(n) = h(1) + h(2) + ...+ h(n-1) + h(n) = 1/1-1/2 + 1/2-1/3 + ...+ 1/(n-1)-1/n + 1/n-1/(n+1)
=1/1 - 1/(n+1)
= n/(n+1),
s(2013) = 2013/2014