1. ## Challenging problem! n roots of a quadratic equation

Hi all

Would appreciate any help!

For each n, let the roots of the quadratic equation $\displaystyle x^2 + (2n+1)x + n^2 = 0$ be $\displaystyle a[n]$ and $\displaystyle b[n]$

Find the value of $\displaystyle 1/(a[3] + 1)(b[3] + 1) + 1/(a[4] + 1)(b[4] + 1) + ... + 1/(a[15] + 1)(b[15] + 1)$

Junior high math can be a killer

2. Originally Posted by Exa
Hi all

Would appreciate any help!

For each n, let the roots of the quadratic equation $\displaystyle x^2 + (2n+1)x + n^2 = 0$ be $\displaystyle a[n]$ and $\displaystyle b[n]$

Find the value of $\displaystyle 1/(a[3] + 1)(b[3] + 1) + 1/(a[4] + 1)(b[4] + 1) + ... + 1/(a[15] + 1)(b[15] + 1)$

Junior high math can be a killer
$\displaystyle x^2 + (2n+1)x + n^2 = 0 \Rightarrow x = \frac{-(2n+1) \pm \sqrt{(2n+1)^2 -4*1*n^2}}{2}$
$\displaystyle \Rightarrow x = \frac{-2n-1 \pm \sqrt{4n^2 + 4n + 1 -4n^2}}{2} = \frac{-2n-1 \pm \sqrt{4n+1}}{2}$

Therefore $\displaystyle a_n = \frac{-2n-1 + \sqrt{4n+1}}{2}, \ b_n = \frac{-2n-1 - \sqrt{4n+1}}{2}$

$\displaystyle (a_n+1)(b_n+1) = (\frac{-2n-1+\sqrt{4n+1}}{2}+1)(\frac{-2n-1-\sqrt{4n+1}}{2}+1)$ $\displaystyle = \frac{-2n+1+\sqrt{4n+1}}{2} \cdot \frac{-2n+1-\sqrt{4n+1}}{2}$

$\displaystyle =\frac{1}{4} (-2n+1 + \sqrt{4n+1})(-2n+1-\sqrt{4n+1}) = \frac{1}{4}((-2n+1)^2 - (4n+1))$ $\displaystyle = \frac{4n^2 -4n + 1 - (4n+1)}{4} = n^2 -2n$

So $\displaystyle (a_n+1)(b_n+1) = n^2-2n$

Can you continue from here?

3. Originally Posted by Exa
Hi all

Would appreciate any help!

For each n, let the roots of the quadratic equation $\displaystyle x^2 + (2n+1)x + n^2 = 0$ be $\displaystyle a[n]$ and $\displaystyle b[n]$

Find the value of $\displaystyle 1/(a[3] + 1)(b[3] + 1) + 1/(a[4] + 1)(b[4] + 1) + ... + 1/(a[15] + 1)(b[15] + 1)$

Junior high math can be a killer
$\displaystyle \frac{1}{(a_n+1)(b_n+1)}=\frac{1}{a_nb_n+(a_n+b_n) +1}$

The product of the roots of $\displaystyle x^2 + (2n+1)x + n^2 = 0$ is $\displaystyle n^2$ and the sum of the roots is $\displaystyle -(2n+1)$, hence:

$\displaystyle \frac{1}{(a_n+1)(b_n+1)}=\frac{1}{n^2-2n}$

CB