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Math Help - Induction

  1. #1
    Member alexgeek's Avatar
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    Red face Induction

    I'm trying to teach myself induction, I get the idea behind it but theres always some weird jump in algebra at some point that throws me in the examples.

    I'm following the example here

    The bit that throws me is how they got from:
     = (n+1)^2 + \frac{1}{6}n(n+1)(2n+1)
    To:
     = \frac{1}{6}(n+1)[6(n+1) + n(2n+1)]

    Doesn't make sense to me how they got rid of the + and the power of 2.

    Thanks.
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  2. #2
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    Smile

    Quote Originally Posted by alexgeek View Post
    I'm trying to teach myself induction, I get the idea behind it but theres always some weird jump in algebra at some point that throws me in the examples.

    I'm following the example here

    The bit that throws me is how they got from:
     = (n+1)^2 + \frac{1}{6}n(n+1)(2n+1)
    To:
     = \frac{1}{6}(n+1)[6(n+1) + n(2n+1)]

    Doesn't make sense to me how they got rid of the + and the power of 2.

    Thanks.
    hi
    they factorized
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by alexgeek View Post
    I'm trying to teach myself induction, I get the idea behind it but theres always some weird jump in algebra at some point that throws me in the examples.

    I'm following the example here

    The bit that throws me is how they got from:
     = (n+1)^2 + \frac{1}{6}n(n+1)(2n+1)
    To:
     = \frac{1}{6}(n+1)[6(n+1) + n(2n+1)]

    Doesn't make sense to me how they got rid of the + and the power of 2.

    Thanks.
    Man, if you got the rest of what they said, and this is the only thing that you have trouble with then consider yourself lucky. The step that "doesn't make sense" to you is simple factoring. Can you see?
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  4. #4
    Member alexgeek's Avatar
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    Oh right ha, just found it hard to spot.
    I'm gunna try and see if I can do one of the examples.
    Thanks
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  5. #5
    Member alexgeek's Avatar
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    I'm trying to prove the sum of an arithmetic series,
    P(n): \sum_{i=1}^{n} a+(n-1)d = \frac{n}{2}(2a+(n-1)d)
    for  a,d,n > 0

    I've proved it for
     Let a = d = 1
    P(1) = \frac{1}{2}(2 + 0) = \frac{2}{2} = 1

    I'm getting stuck on proving
     \sum_{i=1}^{n+1} = a+(n-1)d + \sum_{i=1}^{n}
     \frac{n+1}{2}(2a + ( (n+1) - 1)d ) = a + (n-1)d + \frac{n}{2}(2a + (n-1)d)

    Closest I've got is:
    \frac{2an + 2a}{2} + \frac{n+1}{2}[((n+1)-1)d] = a + (n-1)d + \frac{2an}{2} + \frac{n}{2} ((n-1)d)

    Which gets rid of an and a on both sides, but everything else I do seems to fail leaving me with unequal amounts of d's and n's on each side.

    Anyone got any ideas? thanks
    Last edited by alexgeek; March 2nd 2010 at 12:45 PM.
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  6. #6
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    Smile

    do u mean \sum_{i=1}^{n}a_i or \sum_{i=1}^{n}i.
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  7. #7
    MHF Contributor
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    Hi alex,

    why not do it this way ?

    \sum_{i=1}^n\left(a+[i-1]d\right)=\frac{n}{2}\left(2a+[n-1]d\right)

    \sum_{i=1}^{n+1}\left(a+[i-1]d\right)=\frac{n}{2}\left(2a+[n-1]d\right)+a+nd

    which should equal

    \frac{n+1}{2}\left(2a+nd\right)

    Proof

    \frac{n}{2}\left(2a+[n-1]d\right)+a+nd=\frac{n}{2}(2a+nd)+\frac{n}{2}(-d)+a+nd

    =\frac{n}{2}(2a+nd)-\frac{nd}{2}+\frac{2a}{2}+\frac{2nd}{2}

    =\frac{n}{2}(2a+nd)+\frac{nd}{2}+\frac{2a}{2}

    =\frac{n}{2}(2a+nd)+\frac{1}{2}(2a+nd)=\frac{n+1}{  2}(2a+nd)
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  8. #8
    Super Member Quacky's Avatar
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    Edit: nevermind, this post is not appropriate.
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  9. #9
    Member alexgeek's Avatar
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    Quote Originally Posted by Archie Meade View Post
    Hi alex,

    why not do it this way ?

    \sum_{i=1}^n\left(a+[i-1]d\right)=\frac{n}{2}\left(2a+[n-1]d\right)

    \sum_{i=1}^{n+1}\left(a+[i-1]d\right)=\frac{n}{2}\left(2a+[n-1]d\right)+a+nd

    which should equal

    \frac{n+1}{2}\left(2a+nd\right)

    Proof

    \frac{n}{2}\left(2a+[n-1]d\right)+a+nd=\frac{n}{2}(2a+nd)+\frac{n}{2}(-d)+a+nd

    =\frac{n}{2}(2a+nd)-\frac{nd}{2}+\frac{2a}{2}+\frac{2nd}{2}

    =\frac{n}{2}(2a+nd)+\frac{nd}{2}+\frac{2a}{2}

    =\frac{n}{2}(2a+nd)+\frac{1}{2}(2a+nd)=\frac{n+1}{  2}(2a+nd)

    ahh, so when I was adding the extra term to the right, I forgot to add 1 to n.
    Thanks loads for all that, couldn't find it on the net.
    I'll see if I can do geometric series now
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