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Math Help - Challenging problem! n roots of a quadratic equation

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

    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
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  2. #2
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    Quote Originally Posted by Exa View Post
    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?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Exa View Post
    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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