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Thread: Proving Summation is equal to expression for all positive integers n

  1. #1
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    Thumbs down Proving Summation is equal to expression for all positive integers n

    Prove that ∑ from k =1 to n of 1/(4k^2 - 1) = n/(2n+1) for all positive integers n.

    In words: Prove that the sum (from k =1 to n) of one over (4 times k squared minus one) is equal to n over two times n + 1

    I am beyond beyond beyond lost.
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    Re: Proving Summation is equal to expression for all positive integers n

    Quote Originally Posted by azollner95 View Post
    Prove that ∑ from k =1 to n of 1/(4k^2 - 1) = n/(2n+1) for all positive integers n.
    $\dfrac{1}{4x^2-1}=\dfrac{2^{-1}}{2k-1}-\dfrac{2^{-1}}{2k+1}$
    Then look at collapsing sums.
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    Re: Proving Summation is equal to expression for all positive integers n

    What is a collapsing sum and how do you get there. Lost on a few of the basic steps needed. Is it similar to telescoping series?
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    Re: Proving Summation is equal to expression for all positive integers n

    Quote Originally Posted by azollner95 View Post
    a collapsing sum = to telescoping series?
    Samething

    $\frac{1}{2}\left[ {\left( {\frac{1}{1} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{5}} \right) + \left( {\frac{1}{5} - \frac{1}{7}} \right) \cdots + \left( {\frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right)} \right] = \frac{1}{2}\left[ {1 - \frac{1}{{2n + 1}}} \right]$
    Last edited by Plato; Apr 5th 2017 at 01:13 PM.
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    Re: Proving Summation is equal to expression for all positive integers n

    Quote Originally Posted by azollner95 View Post
    I am beyond beyond beyond lost.
    Are you saying your teacher gave you this problem
    without any teaching?
    Get him/her fired!
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