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Math Help - Induction to prove

  1. #1
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    Question Induction to prove

    Hi, could someone please explain how to do this:

    use induction to prove that
    <br />
\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} <br />

    for all positive integer n.

    [Note: \sum_{i=1}^n i^2 means the sum 1^2+2^2+3^2+...+n^2]

    thank you!!!
    Last edited by lin.13579; August 22nd 2009 at 06:59 PM.
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  2. #2
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    Quote Originally Posted by lin.13579 View Post
    Hi, could someone please explain how to do this:

    use induction to prove that
    <br />
\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} <br />

    for all positive integer n.

    [Note: \sum_{i=1}^n i^2 means the sum 1^2+2^2+3^2+...+n^2]

    thank you!!!
    Amazing what you can find in a few seconds using Google. Key words: proof induction sum squares.

    Read this (one of 747,000 found in 0.28 seconds): http://math.arizona.edu/~kerl/doc/induction.pdf
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Amazing what you can find in a few seconds using Google. Key words: proof induction sum squares.

    Read this (one of 747,000 found in 0.28 seconds): http://math.arizona.edu/~kerl/doc/induction.pdf

    oh, my gosh , thank you , I wrote this eqution for a long time......(LaTex). thanks
    Last edited by mr fantastic; August 22nd 2009 at 07:09 PM. Reason: d --> sh
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  4. #4
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    Mathematical Induction Example 2 --- Sum of Squares


    Problem: For any natural number n , 1^2 + 2^2 + ... + n^2 = n( n + 1 )( 2n + 1 )/6.

    Proof:
    Basis Step: If n = 0, then LHS = 02 = 0, and RHS = 0 * (0 + 1)(2*0 + 1)/6 = 0 .
    Hence LHS = RHS.
    Induction: Assume that for an arbitrary natural number n,
    1^2 + 2^2 + ... + n^2 = n( n + 1 )( 2n + 1 )/6. -------- Induction Hypothesis
    To prove this for n+1, first try to express LHS for n+1 in terms of LHS for n, and use the induction hypothesis.
    Here let us try
    LHS for n + 1 = 1^2 + 2^2 + ... + n^2 + (n + 1)^2 = ( 1^2 + 2^2 + ... + n^2 ) + (n + 1)^2
    Using the induction hypothesis, the last expression can be rewritten as
    n( n + 1 )( 2n + 1 )/6 + (n + 1)^2
    Factoring (n + 1)/6 out, we get
    ( n + 1 )( n( 2n + 1 ) + 6 ( n + 1 ) )/6
    = ( n + 1 )( 2n^2 + 7n + 6 )/6
    = ( n + 1 )( n + 2 )( 2n + 3 )/6 ,
    which is equal to the RHS for n+1.

    Thus LHS = RHS for n+1.

    End of Proof.
    Last edited by quah13579; August 22nd 2009 at 08:13 PM.
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  5. #5
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    Apologies for asking so many questions, but I'm trying to master as many questions as I can, but why does "( 2n + 3 )" represent the next case for (2n+1)? shouldnt it be (2n+2)?
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  6. #6
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    The next case for n is n+1, so the next case for 2n + 1 is 2(n+1) + 1 = 2n + 3
    Last edited by Defunkt; August 27th 2009 at 05:54 AM.
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