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Math Help - Induction - Summation

  1. #1
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    Induction - Summation

    3+6+9+...+3n=\frac{n}{2}(3n+3)
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  2. #2
    Super Member Showcase_22's Avatar
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    When n=1:

    3=\frac{1}{2}(3+3)=3

    Hence true for n=1. Assume true for n=k:

    <br />
3+6+9+...+3k=\frac{k}{2}(3k+3)<br />

    ...and i'll leave it up to you to prove it for n=k+1!
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  3. #3
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    Quote Originally Posted by Showcase_22 View Post

    ...and i'll leave it up to you to prove it for n=k+1!
    Thats where I'm stuck...
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  4. #4
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    Quote Originally Posted by nerdzor View Post
    Thats where I'm stuck...
    What have you tried? How far have you gotten?

    You plugged "k + 1" into the formula:

    . . . . . 3\, +\, 6\, +\, 9\, +\, ...\, +\, 3k\, +\, 3(k\, +\, 1)

    You broke off the "just k" part so you could apply the assumption:

    . . . . . \left(3\, +\, 6\, +\, 9\, +\, ...\, +\, 3k\right)\, +\, 3k\, +\, 3

    . . . . . \frac{k}{2}\left(3k\, +\, 3\right)\, +\, 3k\, +\, 3\, =\, \frac{k(3k\, +\, 3)}{2}\, +\, \frac{6k\, +\, 6}{2}

    . . . . . =\, \frac{3k^2\, +\, 3k\, +\, 6k\, +\, 2}{2}\, =\, \frac{3k^2\, +\, 9k\, +\, 6}{2}

    . . . . . =\, \frac{3(k^2\, +\, 3k\, +\, 2)}{2}\, =\, \frac{3(k\, +\, 2)(k\, +\, 1)}{2}

    . . . . . =\, \frac{3(k\, +\, 1)((k\, +\, 1)\, +\, 1)}{2}

    Where did you go from here?

    Please be complete. Thank you!
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