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Thread: Sequential Proof

  1. #1
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    Sequential Proof

    Prove that the $\displaystyle (2n+1)$th term of the sequence $\displaystyle U_n=n^{2}-1$ is a multiple of $\displaystyle 4$.

    How do I do this?
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  2. #2
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    $\displaystyle U_{2n+1}=(2n+1)^2-1=4n^2+4n+1-1=4(n^2+n)$ So it is a multiple of 4.
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  3. #3
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    EDIT: Nevermind, I just realised what a fool I'm being. I'll try it again soon.
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  4. #4
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    Quote Originally Posted by Viral View Post
    $\displaystyle U_{2n+1}=(2n+1)^2-1=4n^2+1-1$

    That's what I keep getting =\ .
    The expansion of $\displaystyle (2n+1)^2$ should be

    $\displaystyle (2n+1)^2 = 4n^2+4n+1$
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