Sum of roots with a twist

Calculus Level 3

Let $\alpha_{n}, \beta_{n}$ be the distinct roots of the equation $x^{2}+(n+1)x+n^{2} = 0$ . If

$\sum_{n=2}^{2015} \frac{1}{(\alpha_{n}+1)(\beta_{n}+1)}$

can be expressed in the form $\frac{a}{b}$ , where $a$ and $b$ are positive coprime integers, find $b-a$ .

The answer is 1.

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Jake Lai
Feb 28, 2015

By Vieta's, we get that

$(\alpha_{n}+1)(\beta_{n}+1) = \alpha_{n}\beta_{n}+\alpha_{n}+\beta_{n}+1 = n^{2}-(n+1)+1 = n(n-1)$

So, our sum becomes

$\sum_{n=2}^{2015} \frac{1}{n(n-1)} = \sum_{n=2}^{2015} \left( \frac{1}{n-1}-\frac{1}{n} \right) = \frac{1}{2-1}-\frac{1}{2015} = \boxed{\frac{2014}{2015}}$

Nice and easy problem with a perfect solution.

Shubhendra Singh - 6 years, 3 months ago

Since you mentioned that the roots need to be distinct, you might want to show that $\Delta\neq 0~\forall~2\leq n\leq 2015~,~n\in\mathbb{Z}$ .

This can be done by simply setting $\Delta\neq 0$ which gives $n\neq 1,\left(-\frac{1}{3}\right)$ . Since these values of $n$ do not fall within the summation sequence, we proceed as usual.

Prasun Biswas - 6 years, 3 months ago

I thought this part was trivial. It doesn't matter to the answer of the question anyway. Thanks for filling it though; nice exercise on discriminants.

Jake Lai - 6 years, 3 months ago

Sum of roots with a twist

The answer is 1.

1 solution

Discussions for this problem are now closed

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