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Math Help - A simple proof?

  1. #1
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    A simple proof?

    Prove that:
    2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)
    When n > 0

    Getting the n-1 part is easy (the left side of the equation has to have n-1 terms, so the result would be something multiplied by that). I don't know how to get the n, though...
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  2. #2
    Super Member Anonymous1's Avatar
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    Use induction to prove. Show the formula holds for the base case, assume it hold for the n^{th} case and derive the (n+1)^{th} case.

    Why don't you try and let me know where you get stuck.
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  3. #3
    Super Member Anonymous1's Avatar
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    Or wait!

    2 + 4 + 6 + 8 ... + 2(n-1) = 2(1+2+3+...+(n-1)) = 2(\frac{n(n-1)}{2}) = n(n-1)

    You may know the formula to obtain the sum of the numbers from 1 to n. It is \frac{n(n+1)}{2}.

    Here is an example:
    Sum the numbers 1 to 5. Then,

    1+2+3+4+5 = \frac{5\times 6}{2} = 15.
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  4. #4
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    Hm. Gimme a second to type as I'm thinking.

    n = 1
    2(1-1) = 1 * (1-1) = 0

    n = k+1
    2 + 4 + 6 ... + 2k = k^2 + k
    2 + 4 + 6 ... + k = k^2
    2 + 4 + 6 ... + 2(k-1) = k^2 - k
    2 + 4 + 6 ... + 2(k-1) = k * (k-1)

    Oh, wow. Thanks so much!


    EDIT: Or, that second post works too. Heh, either one. Thanks for the two methods, at least.
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  5. #5
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    Quote Originally Posted by BlackBlaze View Post
    Prove that:
    2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)
    When n > 0

    LHS:

    2 + 4 + 6 + 8 +\dots+ 2(n-4)+ 2(n-3) + 2(n-2) + 2(n-1) = S

    Swapping the order of the LHS

     <br />
2(n-1) + 2(n-2)+ 2(n-3) + 2(n-4) +\dots+ 8 + 6+ 4 +2= S<br />

    Adding these

    \underbrace{2n+2n + 2n+\dots+ 2n}_{\text{n-1 times}} = 2S

    2n(n-1) = 2S

    n(n-1) = S

    S = n(n-1)
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  6. #6
    Super Member Bacterius's Avatar
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    Hello,
    here's a proof with the sum symbol (looks cool ).
    What you are trying to prove is that \sum_{k=1}^{n - 1} 2k = n(n - 1). Let's work with this :

    \sum_{k=1}^{n - 1} 2k

    Note that each term is even, we can factorize ! ( 2 + 4 + 6 = 2(1 + 2 + 3)). So :

    \sum_{k=1}^{n - 1} 2k = 2 \left ( \sum_{k=1}^{n - 1} k \right )

    Now you may use the fact that \sum_{k=1}^{n} k = \frac{1}{2}n(n + 1), and your result follows. This is a theorem but it can be proven :

    \sum_{k=1}^{n} k = 1 + 2 + 3 + \cdots + k

    1. Assume n is even.

    Rearranging the terms (putting brackets for better comprehension). This is valid because n is even :

    \sum_{k=1}^{n} k = \left [ 1 + \left ( n - 1 \right ) \right ] + \left [ 2 + \left ( n - 2 \right ) \right ] + \cdots + \frac{n}{2} + n

    The dotted part stops when we reach n - \frac{n}{2} (excluded) because we cannot include that term twice. (*)

    This can be simplified :

    \sum_{k=1}^{n} k = n + n + \cdots + \frac{n}{2} + n

    Using fact (*), we simplify by multiplication :

    \sum_{k=1}^{n} k = \left ( \frac{n}{2} - 1 \right ) n + \frac{n}{2} + n

    Expanding and simplifying :

    \sum_{k=1}^{n} k = \frac{n^2}{2} - n + \frac{n}{2} + n = \frac{n^2}{2} + \frac{n}{2} = \frac{n^2 + n}{2} = \frac{n(n + 1)}{2} = \frac{1}{2} n(n + 1)

    2. Assume n is odd.

    Rearranging the terms (putting brackets for better comprehension). This is valid because n is odd :

    \sum_{k=1}^{n} k = \left [ 1 + n \right ] + \left [ 2 + \left ( n - 1 \right ) \right ] + \cdots + \frac{n + 1}{2}

    The dotted part stops when we reach n - \frac{n + 1}{2} (excluded) because if we included it we would include two terms twice! (*)

    This can be simplified :

    \sum_{k=1}^{n} k = (n + 1) + (n + 1) + \cdots + \frac{n + 1}{2}

    Using fact (*), we simplify by multiplication :

    \sum_{k=1}^{n} k = \left ( \frac{n + 1}{2} - 1 \right )(n + 1) + \frac{n + 1}{2}

    Expanding and simplifying (using the trick that \frac{n + 1}{2} - 1 = \frac{n - 1}{2}) :

    \sum_{k=1}^{n} k = \left ( \frac{n + 1}{2} - 1 \right )(n + 1) + \frac{n + 1}{2} = \frac{(n - 1)(n + 1)}{2} + \frac{n + 1}{2} = \ \frac{(n - 1)(n + 1) + (n + 1)}{2} = \frac{(n + 1)(n - 1 + 1)}{2} = \frac{1}{2} n(n + 1)



    Using the theorem we just proved (the proof is correct, just a bit messed up, needs better editing), we get that :

    \sum_{k=1}^{n - 1} 2k = 2 \sum_{k=1}^{n - 1} k = 2 \times \left ( \frac{1}{2} (n - 1)(n - 1 + 1) \right ) = n(n - 1)

    And we can conclude.

    Does it make sense ?

    EDIT : I just saw Pickslides' proof. Mine seems a bit redundant now
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  7. #7
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    Quote Originally Posted by BlackBlaze View Post
    Prove that:
    2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)
    When n > 0

    Getting the n-1 part is easy (the left side of the equation has to have n-1 terms, so the result would be something multiplied by that). I don't know how to get the n, though...
    Well, the sum of n-1 number is n-1 times their average. Since this is an arithmetic sequence, the average of all the numbers is just the average of the smallest and largest: \frac{2+ 2(n-1)}{2}= 1+ (n- 1)= n.
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