MHB Roots of Equations & Sum of Inverses: $a=1,2,3,\dots,2011$

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The discussion focuses on finding the sum of the inverses of the roots of the quadratic equations defined by $x^2 - 2x - a^2 - a = 0$ for $a$ ranging from 1 to 2011. Participants highlight issues with the last step of the calculation, indicating that it was incorrect initially. A correction is proposed to accurately compute the sum $\sum_{n=1}^{2011}(\dfrac{1}{\alpha_n}+\dfrac{1}{\beta_n})$. The roots of the equations are denoted as $(\alpha_n, \beta_n)$ for each value of $a$. The discussion emphasizes the importance of verifying calculations in mathematical problem-solving.
Albert1
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$a=1,2,3,4,5,------2011$, the roots of the equations $x^2-2x-a^2-a=0,$ are :
$(\alpha_1,\beta_1),(\alpha_2,\beta_2),----------,(\alpha_{2011},\beta_{2011})$ respectively
please find :
$\sum_{n=1}^{2011}(\dfrac{1}{\alpha_n}+\dfrac {1}{\beta_n})$
 
Last edited:
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Albert said:
$a=1,2,3,4,5,------2011$, the roots of the equations $x^2-2x-a^2-a=0,$ are :
$(\alpha_1,\beta_1),(\alpha_2,\beta_2),----------,(\alpha_{2011},\beta_{2011})$ respectively
please find :
$\sum_{n=1}^{2011}(\dfrac{1}{\alpha_n}+\dfrac {1}{\beta_n})$

$(\alpha_n,\beta_n)$ are roots of equation $x^2-2x-n^2-n=0$
so
$\alpha_n+\beta_n= 2$
$(\alpha_n\beta_n)=- (n^2+n) = -n(n+1)$
hence
$\dfrac{1}{\alpha_n}+\dfrac{1}{\beta_n}$
= $\dfrac{\alpha_n+\beta_n}{\alpha_n\beta_n}$
= $\dfrac{-2}{n^2+n}$
= -2$(\dfrac{1}{n}-\dfrac{1}{n+1})$

now the above is telesopic sum and the sum of 2011 terms is
-2$(1 -\dfrac{1}{2012})$
=$- \dfrac{4013}{2012}$

The last step calculation was not correct. it is corrected to $- \dfrac{2011}{1006}$ based on comment
 
Last edited:
kaliprasad said:
$(\alpha_n,\beta_n)$ are roots of equation $x^2-2x-n^2-n=0$
so
$\alpha_n+\beta_n= 2$
$(\alpha_n\beta_n)=- (n^2+n) = -n(n+1)$
hence
$\dfrac{1}{\alpha_n}+\dfrac{1}{\beta_n}$
= $\dfrac{\alpha_n+\beta_n}{\alpha_n\beta_n}$
= $\dfrac{-2}{n^2+n}$
= -2$(\dfrac{1}{n}-\dfrac{1}{n+1})$

now the above is telesopic sum and the sum of 2011 terms is
-2$(1 -\dfrac{1}{2012})$
=$- \dfrac{4013}{2012}$
the last step ,calculation not correct
 
Albert said:
the last step ,calculation not correct

corrected the last step in the calculation
 
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