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Math Help - Induction for Triangular numbers

  1. #1
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    Induction for Triangular numbers

    I have tried and tried, with no success. Instructor is useless, book is far more useless, there's nothing worse than online math class. Can anyone help me answer this problem? I am trying to do my homework, I've done one number right.

    I need to answer 11 more, this is the 2nd of them.

    This is discrete math, Induction.

    This is the question:

    Alumberjack has 4n + 110 logs in a pile consisting of n layers.
    Each layer has two more logs than the layer directly above
    it. If the top layer has six logs, how many layers are there?

    And i attached an image of an example that is somewhat similar to what I am supposed to do.

    Induction for Triangular numbers-4.5.1.png
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Induction for Triangular numbers

    The number of logs is given by:

    2\sum_{k=1}^n(k+2)=4n+110

    Can you solve for n?
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  3. #3
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    Re: Induction for Triangular numbers

    I'm sorry I am really lost. (on the verge of paying an online tutor to help me pass this class). Can you help me understand how you have come up with this formula?

    I am guessing that n can be any number?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Induction for Triangular numbers

    We are told the top layer of logs contains 6 logs, and that each layer has two logs more than the layer above it. So, the number of logs is:

    S=6+8+10+\cdots

    Now, writing these terms as a function of the layer number, counting from the top, we have:

    S=(2(1)+4)+(2(2)+4)+(2(3)+4)+\cdots+(2(n)+4)

    We may factor a 2 out as follows:

    S=2\left((1+2)+(2+2)+(3+2)+\cdots+(n+2) \right)

    and so using sigma notation, we may write:

    S=2\sum_{k=1}^n(k+2)

    Now, we are also told that:

    S=4n+110

    and so, we have:

    2\sum_{k=1}^n(k+2)=4n+110

    To evaluate the sum, let's go back to:

    S=2\left((1+2)+(2+2)+(3+2)+\cdots+(n+2) \right)

    which we may write as:

    \frac{S}{2}=3+4+5+\cdots+(n+2)

    \frac{S}{2}=(n+2)+(n+1)+n+\cdots+3

    Adding the two equations, we find:

    S=n(n+5)

    Do you see how we have n + 5 a total of n times?

    So now we have:

    n(n+5)=4n+110

    This is a quadratic that factors nicely, but does have a negative root you will need to discard. Does this make more sense?
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  5. #5
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    Re: Induction for Triangular numbers

    Hello, jpab29!

    This is not an Induction problem.
    The problem involves an Arithmetic Series.


    \text{A lumberjack has }4n + 110\text{ logs in a pile consisting of }n\text{ layers.}
    \text{Each layer has two more logs than the layer directly above it.}
    \text{If the top layer has six logs, how many layers are there?}

    The sum of an arithmetic series is: . S_n \;=\;\tfrac{n}{2}\big[2a + (n-1)d\big]

    . . where: . \begin{Bmatrix}n &=& \text{number of terms} \\ a &=& \text{first term} \\ d &=& \text{common difference} \end{Bmatrix}


    We are given: . \begin{Bmatrix}S_n &=& 4n+110 \\ a &=& 6 \\ d &=& 2\end{Bmatrix}

    We have: . \tfrac{n}{2}\big[2(6) + (n-1)2\big] \:=\:4n+110

    n . . . . . . . . . \tfrac{n}{2}\big[12 + 2n - 2\big] \:=\:4n + 110

    . . . . . . . . . . . . . \tfrac{n}{2}(2n + 10) \:=\:4n+110

    . . . . . . . . . . . . . . . n^2 + 5n \:=\:4n + 110

    n . . . . . . . . . . n^2 + n - 110 \:=\:0

    . . . . . . . . . (n-10)(n+11) \:=\:0


    Therefore: . n \:=\:10
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  6. #6
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    Re: Induction for Triangular numbers

    Thank you! But I lost track at this part:

    Induction for Triangular numbers-abc.png
    Last edited by jpab29; July 10th 2013 at 11:22 PM.
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  7. #7
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    Re: Induction for Triangular numbers

    Thank you so much! I will try to analyze it!
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  8. #8
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    Re: Induction for Triangular numbers

    I found a youtube video with the formula that you just used. Thanks so much!

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  9. #9
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    Re: Induction for Triangular numbers

    With the induction formula that you came up with, will it also end up with Soroban's solution?
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  10. #10
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    Re: Induction for Triangular numbers

    Quote Originally Posted by jpab29 View Post
    With the induction formula that you came up with, will it also end up with Soroban's solution?
    I would call it a summation formula instead, but yes, you can see it leads to the same quadratic in n.
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  11. #11
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    Re: Induction for Triangular numbers

    Okay. I understood the arithmetic series more than the summation formula though.
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  12. #12
    MHF Contributor MarkFL's Avatar
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    Re: Induction for Triangular numbers

    You have indicated that you had trouble understanding this part of my post:

    \frac{S}{2}=3+4+5+\cdots+(n+2)

    \frac{S}{2}=(n+2)+(n+1)+n+\cdots+3

    Adding the two equations, we find:

    S=n(n+5)
    If we add all the corresponding terms in both equations, we get:

    \frac{S}{2}+\frac{S}{2}=(3+(n+2))+(4+(n+1))+(5+n)+  \cdots+((n+2)+3)

    S=(n+5)+(n+5)+(n+5)+\cdots+(n+5)

    On the right, there are n terms, thus:

    S=n(n+5)

    And so:

    n(n+5)=4n+110

    n^2+5n=4n+110

    n^2+n-110=0

    (n-10)(n+11)=0

    Discarding the negative root, we are left with:

    n=10

    Does this make more sense?
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