Challenging problem! n roots of a quadratic equation

**Exa** · September 26th 2009, 11:23 PM

Hi all

Would appreciate any help!

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

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

Junior high math can be a killer

**Defunkt** · September 27th 2009, 01:34 AM

Originally Posted by Exa

Hi all

Would appreciate any help!

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

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

Junior high math can be a killer

$x^2 + (2n+1)x + n^2 = 0 \Rightarrow x = \frac{-(2n+1) \pm \sqrt{(2n+1)^2 -4*1*n^2}}{2}$
$\Rightarrow x = \frac{-2n-1 \pm \sqrt{4n^2 + 4n + 1 -4n^2}}{2} = \frac{-2n-1 \pm \sqrt{4n+1}}{2}$

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

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

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

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

Can you continue from here?

**CaptainBlack** · September 27th 2009, 02:06 AM

Originally Posted by Exa

Hi all

Would appreciate any help!

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

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

Junior high math can be a killer

$\frac{1}{(a_n+1)(b_n+1)}=\frac{1}{a_nb_n+(a_n+b_n) +1}$

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

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

CB

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