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Math Help - Proving Identities

  1. #1
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    Proving Identities

    How do I show that the sum from {i=1} to {n} (2i-1) is equal to n^2 for all natural numbers n?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Try induction!
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  3. #3
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    I don't know what that is. XD
    We haven't learned it yet.
    When I look at the formulas and rules, I find nothing that helps at all.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Well I can give you the identity \sum_{j=1}^nj=\frac{n(n+1)}{2}. See if you can solve the problem using this, and then if you wish we will look at how to prove it.
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  5. #5
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    Alright. Now i've gotten the original sum down to this:

    2* sum from {i=1} to {n} i = (n(n+1))/2 - sum from {i=1} to {n} 1 = n.

    Now I think I have to prove it, but I don't know how.
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Good. Be careful with your notation - be as clear as possible, don't put more than one equal sign in your expression!

    So you have \sum_{j=1}^n(2j-1)=2\left\{n(n+1)/2\right\}-n=n^2+n-n=n^2.

    Now, to prove that 1+2+...+n=n(n+1)/2, consider

    S=1+2+...+n

    and add it to the same sum, written backwards :

    S=n+(n-1)+..+1.

    Add them vertically, term by term. What do you get?
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  7. #7
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    I'm confused as to what you are asking.
    I have to prove that all of that equals n^2, for all natural numbers n
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    Yes; I have showed you how to do that, but I've used an identity which we have not proved, namely 1+2+...+n=n(n+1)/2. I was suggesting a proof of the latter identity.
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  9. #9
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    Oh ok. I have one question. Since the problem has 2 in front of the sigma notation in the first term, could i cancel the two out to get sum from {i=1} to {n} n^2 + n - sum from {i=1} to {n} n?
    And then I could cancel one n out, which would give me
    sum from {i=1} to {n} n^2?
    I think that is how you prove it, but I cannot be sure
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  10. #10
    MHF Contributor Bruno J.'s Avatar
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    Yes, that is what you have to do. Read my post above :

    Quote Originally Posted by Bruno J. View Post
    So you have \sum_{j=1}^n(2j-1)=2\left\{n(n+1)/2\right\}-n=n^2+n-n=n^2.
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  11. #11
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    Oh ok. I looked at that equation and I just didn't think. Thank you though! You pointed me in the right direction, and gave me the answer, but I didn't realize you gave me the answer, and I got it myself.
    Thanks!
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