Evaluate $\frac{1}{m_1} +\frac{1}{n_1} +\frac{1}{m_2} +\frac{1}{n_2}+...+\frac{1}{m_{2011}} +\frac{1}{n_{2011}}$ When $a=1,2,3,...,2010,2011$ the roots of the equation $x^2 -2x-a^2-a=0$ are $(m_1,n_1 ), (m_2,n_2 ), (m_3,n_ 3),..., (m_{2010},n_{2010} ), (m_{2011},n_{2011 }) $ respectively. Evaluate $\frac{1}{m_1} +\frac{1}{n_1} +\frac{1}{m_2} +\frac{1}{n_2}+...+\frac{1}{m_{2010}} +\frac{1}{n_{2010}} +\frac{1}{m_{2011}} +\frac{1}{n_{2011}}$
**My attempt **$$$$
$ \frac{1}{m_a} +\frac{1}{n_a}$can be manipulated to give
$$\frac{m_a+n_a}{m_an_a}$$then w.l.o.g $m_a\ge n_a$
And since$$m_a=2\frac{2+2\sqrt{a^2+a+1}}{2}$$
And $$n_a= 2\frac{2-2\sqrt{a^2+a+1} }{2}$$
We get
$$\frac{m_a+n_a}{m_an_a}=\frac{-2}{a(a+1)}$$
But I dont know what to do next,
Suggestions and solutions would be appreciated
Taken from the 2011 IWYMIC

You don’t need to compute the roots $n_a, m_a$. Just note that: $\dfrac{1}{n_a}+\dfrac{1}{m_a}=\dfrac{n_a+m_a}{n_am_a}=\dfrac{S_a}{P_a}$ where $S_a$ and $P_a$ denote the sum and the product of those two roots. And:
$\dfrac{S_a}{P_a}=\dfrac{-b}{c}=-\dfrac{2}{a(a+1)}=2\left(\dfrac{1}{a+1}-\dfrac1a\right)$. Finally:
$$\sum_{a=1}^{2011}{\dfrac{S_a}{P_a}}=2\sum_{a=1}^{2011}{\left(\dfrac{1}{a+1}-\dfrac1a\right)}=2\left(\dfrac{1}{2012}-\dfrac11\right)$$
