Reciprocal sum of roots

Level pending

For $i=1,2...,2014$ let $\alpha_i$ and $\beta_i$ be the roots of

$x^2-2x-i^2-i=0$ .

The value of

$\displaystyle\sum_{i=1}^{2014}\frac{1}{\alpha_i}+\frac{1}{\beta_i}$

can be expressed as $-\frac{a}{b}$ , where a and b are coprime positive integers.What is the value of b-2014?

The answer is 1.

1 solution

Bogdan Simeonov
Jan 3, 2014

By Vieta's formulas we get that $\alpha_i+\beta_i=2$ .

The sum $\displaystyle\sum_{i=1}^{2014}\frac{1}{\alpha_i}+\frac{1}{\beta_i}$ = $\displaystyle\sum_{i=1}^{2014}\frac{\alpha_i+\beta_i}{\alpha_i.\beta_i}$ . Also from Vieta, we get that $\alpha_i.\beta_i=-i^2-i=-i(i+1)$ .

So we get for our initial sum that

$\displaystyle\sum_{i=1}^{2014}\frac{1}{\alpha_i}+\frac{1}{\beta_i}=(-2).\displaystyle\sum_{i=1}^{2014}\frac{1}{i(i+1)}=(-2).\frac{2014}{2015}=-\frac{4028}{2015}$ .

So b=2015 and $b-2014=\boxed1$

How did you get that $\displaystyle \sum_{i=1}^{n} \; \dfrac{1}{i(i+1)} = \dfrac{n}{n+1}?$

Can you prove it beyond "deduction'?

Guilherme Dela Corte - 7 years, 5 months ago

We use that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ .

Then everything cancels except $1-\frac{1}{2015}=\frac{2014}{2015}$

Bogdan Simeonov - 7 years, 5 months ago

Reciprocal sum of roots

The answer is 1.

1 solution

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