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Math Help - [SOLVED] prove by mathematical induction...it's not true

  1. #1
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    [SOLVED] prove by mathematical induction...it's not true

    Prove by mathematical induction that 2+5+8+...+(3k-1) = 1/2n(3n+1)

    k^2 = 2+5+8+...+(3k-1)
    (k+1)^2 = 2+5+8+...+ 3(k+1) + [3(k+1)-1]
    (k+1)^2 = 2+5+8+...+ 3(k+1) + 3k + 2
    (k+1)^2 = k^2+3k+2
    k^2 + 2k +1 =/ k^2 +3k +2

    Did I do something wrong or is the problem not true?

    Last edited by ninobrn99; May 15th 2009 at 11:21 AM.
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  2. #2
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    I assume you mean 2+5+8+....+(3n-1)=\frac{1}{2}n(3n+1)

    Add 3k+2 to both sides:

    2+5+8+.....+(3k-1)+(3k+2)=\frac{1}{2}k(3k+1)+(3k+2)

    2+5+8+....+(3k-1)+(3k+2)=\frac{1}{2}(k+1)(3(k+1)+1)

    It is shown.

    You can put in the usual details.
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  3. #3
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    Quote Originally Posted by galactus View Post
    I assume you mean 2+5+8+....+(3n-1)=\frac{1}{2}n(3n+1)

    Add 3k+2 to both sides:

    2+5+8+.....+(3k-1)+(3k+2)=\frac{1}{2}k(3k+1)+(3k+2)

    2+5+8+....+(3k-1)+(3k+2)=\frac{1}{2}(k+1)(3(k+1)+1)

    It is shown.

    You can put in the usual details.
    is the right side of the equation:
    1.5k^2 + 3.5k +2??
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  4. #4
    Eater of Worlds
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    Yes, it is. But, to show induction nicely, you may want to keep it in the k+1 form I showed you.
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  5. #5
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    omg, im so lost! does 2+5+8+...(3k-1) = k^2?

    so should it be:
    K^2+3k+2 = (3k+4)(k+1)/2

    This book makes no freaking sense to me!!
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