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Math Help - Quick mathematical induction question

  1. #1
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    Quick mathematical induction question

    The problem


    n
    Σi^2 = n(n + 1)(2n + 1)/6
    i=1

    The base case is correct. Gona get to the meat of the problem

    P(k) -------> P(k +1)
    1^2 + 2^2 + 3^2 ......+k^2 = k(k +1)(2k+ 1)/6

    Now showing k +1
    1^2 + 2^2 + 3^2 ......+k^2 + (K +1)^2 = k+1((k +1)+1)((2k+ 1)+1)/6

    RHS = (2k^3 + 7k^2 + 10k + 6)/6

    LHS =
    1^2 + 2^2 + 3^2 ......+k(k +1)(2k+ 1)/6 + (K +1)^2


    Before I go on was I right in mulitplyong everything out with k +1 on the RHS so i could math it up with the LHS? in bold


    Thank you!!


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  2. #2
    Senior Member BAdhi's Avatar
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    It's bit unclear what you've mentioned

    what have you done before the last line?
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  3. #3
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    Quote Originally Posted by BAdhi View Post
    It's bit unclear what you've mentioned

    what have you done before the last line?
    Thats the right hand sides multiplied together when I replaced the k with k +1
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  4. #4
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    Quote Originally Posted by Thetheorycase View Post
    LHS =
    1^2 + 2^2 + 3^2 ......+k(k +1)(2k+ 1)/6 + (K +1)^2
    You have left out an important equals sign! You mean, of course,
    1^2+ 2^2+ 3^2+ ...+ k^2+ (k+1)^2= k(k+1)(2k+1)/6+ (k+1)^2

    Imstead of multiplying that out, factoring k+ 1 out of the right side you get (k+1)(k(2k+1)/6+ k+1)
    k(2k+1)/6+ k+1= [k(2k+1)+ 6(k+1)]/6= [2k^2+ 7k+ 6]/6
    can you factor that numerator?
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  5. #5
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    Quote Originally Posted by Thetheorycase View Post
    The problem


    n
    Σi^2 = n(n + 1)(2n + 1)/6
    i=1

    The base case is correct. Gona get to the meat of the problem

    P(k) -------> P(k +1)
    1^2 + 2^2 + 3^2 ......+k^2 = k(k +1)(2k+ 1)/6

    Now showing k +1
    1^2 + 2^2 + 3^2 ......+k^2 + (K +1)^2 = k+1((k +1)+1)((2k+ 1)+1)/6

    RHS = (2k^3 + 7k^2 + 10k + 6)/6

    LHS =
    1^2 + 2^2 + 3^2 ......+k(k +1)(2k+ 1)/6 + (K +1)^2


    Before I go on was I right in mulitplyong everything out with k +1 on the RHS so i could math it up with the LHS? in bold


    Thank you!!



    P(k+1)

    It looks like you multiplied out

    \displaystyle\frac{\left(k^2+2k+2\right)\left(2k+3  )}{6}=\frac{2k^3+7k^2+10k+6}{6}

    which means you haven't correctly multiplied out the RHS, which should have read

    \displaystyle\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}=\frac{(k+1)(k+2)(2k+3)}{6}=\frac{\left(k^2+3k  +2\right)(2k+3)}{6}

    You don't have to multiply the factors out

    You can keep RHS= \displaystyle\frac{(k+1)(k+2)(2k+3)}{6}

    Instead you can show the LHS is this

    \displaystyle\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{  k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}

    Both terms have a common factor of \displaystyle\frac{k+1}{6}

    \displaystyle\frac{k+1}{6}\;[{k(2k+1)+6(k+1)]=\frac{k+1}{6}\left(2k^2+7k+6\right)

    and you can finish up by factoring 2k^2+7k+6
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