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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 Kevin

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  • $\begingroup$ Please see math.meta.stackexchange.com/questions/5020 $\endgroup$ – Lord Shark the Unknown Feb 23 at 12:01
  • $\begingroup$ I've already tried using series, but i have not been able to solve it $\endgroup$ – kenith Feb 23 at 12:04
  • $\begingroup$ Any other suggestions, are welcome $\endgroup$ – kenith Feb 23 at 12:04
  • $\begingroup$ My first thought is to "complete the square". The given equation, $x^2- 2x- a^2- a$ is the same as $x^2- 2x= a^2+ a$. We can "complete the square" on the left by adding 1 to both sides: $x^2- 2x+ 1= (x- 1)^2= a^2+ a+ 1$. So $x- 1= \pm\sqrt{a^2+ a}$ and $x= 1\pm\sqrt{a^2+ a}$. For so So when a=1, the roots are $a_1= 1+ \sqrt{2}$ and $b_1= 1- \sqrt{2}$. $\frac{1}{a_1}+ \r $\endgroup$ – user247327 Feb 23 at 12:08
  • $\begingroup$ You found a pattern in the expressions $1/a_i + 1/b_i$. Can you also find a pattern in the partial sums you get by adding them together one by one? $\endgroup$ – FredH Feb 23 at 12:15
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$$\sum_{k=1}^{2011}\left(\frac{1}{a_k}+\frac{1}{b_k}\right)=\sum_{k=1}^{2011}\frac{a_k+b_k}{a_kb_k}=\sum_{k=1}^{2011}\frac{2}{-k(k+1)}=$$ $$=-2\sum_{k=1}^{2011}\left(\frac{1}{k}-\frac{1}{k+1}\right)=-2\left(1-\frac{1}{2012}\right)=-\frac{2011}{1006}.$$

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    $\begingroup$ You're a genius $\endgroup$ – kenith Feb 23 at 12:29
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    $\begingroup$ Thank you very much and I wish I had you as a maths teacher $\endgroup$ – kenith Feb 23 at 12:30
  • $\begingroup$ @kenith You are welcome! Good luck! $\endgroup$ – Michael Rozenberg Feb 23 at 12:32
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    $\begingroup$ can I get in touch with you again if I have further problems? $\endgroup$ – kenith Feb 23 at 12:35
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    $\begingroup$ thank you very much $\endgroup$ – kenith Feb 23 at 12:38
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My first thought is to "complete the square". The given equation, $x^2- 2x- a^2- a$ is the same as $x^2- 2x= a^2+ a$. We can "complete the square" on the left by adding 1 to both sides: $x^2- 2x+ 1= (x- 1)^2= a^2+ a+ 1$. So $x- 1= \pm\sqrt{a^2+ a}$ and $x= 1\pm\sqrt{a^2+ a}$. Now pair the reciprocals: $\frac{1}{a_i}+ \frac{1}{b_i}= \frac{1}{1+ \sqrt{a^2+ a}}+ \frac{1- \sqrt{a^2+ a}}= \frac{1- \sqrt{a^2+ a}}{1- a^2- a}+ $$ \frac{1+ \sqrt{a^2+ a}}{1- a^2- a}= \frac{2}{1- a^2- a}$. The sum of reciprocals reduces to $-2- \frac{2}{5}- \frac{2}{11}- \frac{2}{19}- ...$.

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  • $\begingroup$ Completing the square is fine, but it should be $x-1=\pm \sqrt{a^2+a+1}$. And then: $\frac1{a_i}+\frac1{b_i}=-\frac2{a^2+a}$, so it is a round way. Easier to use Vieta's: $\frac1{a_i}+\frac1{b_i}=\frac{a_i+b_i}{a_ib_i}$. Luck. $\endgroup$ – farruhota Feb 23 at 12:52

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