Sum of reciprocals of roots for x² - 2x - a² - a = 0

[Albert1]

$a=1,2,3,4,5,------2011$, 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 :
$\sum_{n=1}^{2011}(\dfrac{1}{\alpha_n}+\dfrac {1}{\beta_n})$

 

[kaliprasad]

$a=1,2,3,4,5,------2011$, 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 :
$\sum_{n=1}^{2011}(\dfrac{1}{\alpha_n}+\dfrac {1}{\beta_n})$

Spoiler

$(\alpha_n,\beta_n)$ 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

 

[Albert1]

Spoiler

$(\alpha_n,\beta_n)$ 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

 

[kaliprasad]

the last step ,calculation not correct

corrected the last step in the calculation