# Induction for Triangular numbers

• Jul 10th 2013, 09:55 PM
jpab29
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. :)

Attachment 28777
• Jul 10th 2013, 10:32 PM
MarkFL
Re: Induction for Triangular numbers
The number of logs is given by:

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

Can you solve for $\displaystyle n$?
• Jul 10th 2013, 10:43 PM
jpab29
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?
• Jul 10th 2013, 11:09 PM
MarkFL
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:

$\displaystyle S=6+8+10+\cdots$

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

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

We may factor a 2 out as follows:

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

and so using sigma notation, we may write:

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

Now, we are also told that:

$\displaystyle S=4n+110$

and so, we have:

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

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

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

which we may write as:

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

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

Adding the two equations, we find:

$\displaystyle S=n(n+5)$

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

So now we have:

$\displaystyle n(n+5)=4n+110$

This is a quadratic that factors nicely, but does have a negative root you will need to discard. Does this make more sense?
• Jul 10th 2013, 11:13 PM
Soroban
Re: Induction for Triangular numbers
Hello, jpab29!

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

Quote:

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

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

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

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

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

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

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

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

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

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

Therefore: .$\displaystyle n \:=\:10$
• Jul 10th 2013, 11:19 PM
jpab29
Re: Induction for Triangular numbers
Thank you! But I lost track at this part:

Attachment 28778
• Jul 10th 2013, 11:24 PM
jpab29
Re: Induction for Triangular numbers
Thank you so much! I will try to analyze it!
• Jul 10th 2013, 11:36 PM
jpab29
Re: Induction for Triangular numbers
I found a youtube video with the formula that you just used. Thanks so much!

Finding the Sum of a Finite Arithmetic Series - YouTube
• Jul 10th 2013, 11:37 PM
jpab29
Re: Induction for Triangular numbers
With the induction formula that you came up with, will it also end up with Soroban's solution?
• Jul 10th 2013, 11:46 PM
MarkFL
Re: Induction for Triangular numbers
Quote:

Originally Posted by jpab29
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 $\displaystyle n$.
• Jul 10th 2013, 11:47 PM
jpab29
Re: Induction for Triangular numbers
Okay. I understood the arithmetic series more than the summation formula though.
• Jul 11th 2013, 12:05 AM
MarkFL
Re: Induction for Triangular numbers
You have indicated that you had trouble understanding this part of my post:

Quote:

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

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

Adding the two equations, we find:

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

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

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

On the right, there are $\displaystyle n$ terms, thus:

$\displaystyle S=n(n+5)$

And so:

$\displaystyle n(n+5)=4n+110$

$\displaystyle n^2+5n=4n+110$

$\displaystyle n^2+n-110=0$

$\displaystyle (n-10)(n+11)=0$

Discarding the negative root, we are left with:

$\displaystyle n=10$

Does this make more sense?