<s>[INST] Question from the 2011 IMC (International Mathematics Competition) Key Stage III paper, about the evaluation of a quadratic equation 
When $a=1, 2, 3, ..., 2010, 2011$, the roots of the equation $x^2-2x-a^2-a=0$ are $(a_1, b_1), (a_2, b_2), (a_3, b_3),\cdots, (a_{2010}, b_{2010}), (a_{2011}, b_{2011})$ respectively. Evaluate:
$$
\frac{1}{a_1} + \frac{1}{b_1} + \frac{1}{a_1} + \frac{1}{b_2} + \frac{1}{a_3} + \frac{1}{b_3} +\cdots + \frac{1}{a_{2010}} + \frac{1}{b_{2010}} + \frac{1}{a_{2011}} + \frac{1}{b_{2011}}
$$

I tried solving this question, with the use of the quadratic equation.
Using the quadratic equation, I concluded that $a_1=\frac{2+\sqrt{12}}{2}=1+\sqrt{3}$ and that $b_1=\frac{2-\sqrt{12}}{2}=1-\sqrt{3}$. The reason that I concluded to this is because the quadratic equation states:
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
So I substituted $a$ with $1$ ($1$ is multiplying $x^2$, in the original equation), $b$ with $-2$ ($x$ is getting multiplied by $-2$ in the original equation) and $c$ with $(-a^2-a)$, as they are the only ones which are not getting directly multiplied by $x$ in the original equation.
Hence, I subsequently worked out that $\frac{1}{a_1} + \frac{1}{b_1} = \frac{2}{-2}=-1$
Continuing to do the same thing I worked out $\frac{1}{a_2} + \frac{1}{b_2}= \frac{2}{-6} = -\frac{1}{3}$ and $\frac{1}{a3} + \frac{1}{b_3}= \frac{2}{-12}=-\frac{1}{6}$ and $\frac{1}{a_4} + \frac{1}{b_4} = \frac{2}{-20}=-\frac{1}{10}$ and $\frac{1}{a_5} + \frac{1}{b5} = \frac{2}{-30} =-\frac{1}{15}$.
I subsequently realised that a pattern was emerging, the denominator, each time is getting increased by the degree of $n$ at which $a$ and $b$ are (for instance $\frac{1}{a_4} + \frac{1}{b_4}= -\frac{1}{6+4}$)
Had I not been dealing with fractions, I would have solved it using arithmetic progressions, but unfortunately that is not possible.
I can think of no other way of finishing off my thoughts, nor any other way to solve this problem. Can you please help me? Can you please tell me if there is any method of finishing off my thoughts and if there isn't, can you please suggest a method of solving the problem
Thank you in advance
 [/INST] $$\sum_{k=1}^{2011}\left(\frac{1}{a_k}+\frac{1}{b_k}\right)=\sum_{k=1}^{2011}\frac{a_k+b_k}{a_kb_k}=\sum_{k=1}^{2011}\frac{2}{-k(k+1)}=$$
$$=-2\sum_{k=1}^{2011}\left(\frac{1}{k}-\frac{1}{k+1}\right)=-2\left(1-\frac{1}{2012}\right)=-\frac{2011}{1006}.$$
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