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Math Help - Mathematical induction?

  1. #1
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    Mathematical induction?

    Use mathematical induction to prove that each proposition is valid for all positive integral values of n


    1^2+3^2+5^2+...+(2n-1)^2=n(2n-1)(2n+1)/3 This is what I did

    (2k+1)^2

    k(2k-1)(2k+1)/3


    k(2k-1)(2k+1)/3 + (2k+1)^2/1

    k(2k-1)(2k+1)/3 + 3(2k+1)^2/3

    k(2k-1)(2k+1)/3 + 3(2k+1)(2k+1)/3 Now I am kind of stuck


    Basically what I need is a formula that when n=k+1 is applied to n(2n-1)(2n+1)/3

    it will give me the same result but I need to prove it.
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  2. #2
    Senior Member BAdhi's Avatar
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    you've almost come to the end. All you have to do is simplify the final expression (where you said you are stuck) which eventually gives you \frac{(k+1)(2k+1)(2k+3)}{3}
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  3. #3
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    How would you simplify it do you just use distributive property because the (2k+1) on the left side is confusing me.
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  4. #4
    Senior Member BAdhi's Avatar
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    why, (2k+1) is common to both sides, take it out and simplify
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