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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. :)

Attachment 28777

Re: Induction for Triangular numbers

The number of logs is given by:

Can you solve for ?

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?

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:

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

We may factor a 2 out as follows:

and so using sigma notation, we may write:

Now, we are also told that:

and so, we have:

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

which we may write as:

Adding the two equations, we find:

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

So now we have:

This is a quadratic that factors nicely, but does have a negative root you will need to discard. Does this make more sense?

Re: Induction for Triangular numbers

Hello, jpab29!

This is not an Induction problem.

The problem involves an Arithmetic Series.

The sum of an arithmetic series is: .

. . where: .

We are given: .

We have: .

n . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . . . .

n . . . . . . . . . .

. . . . . . . . .

Therefore: .

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Re: Induction for Triangular numbers

Thank you! But I lost track at this part:

Attachment 28778

Re: Induction for Triangular numbers

Thank you so much! I will try to analyze it!

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

Re: Induction for Triangular numbers

With the induction formula that you came up with, will it also end up with Soroban's solution?

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 .

Re: Induction for Triangular numbers

Okay. I understood the arithmetic series more than the summation formula though.

Re: Induction for Triangular numbers

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

If we add all the corresponding terms in both equations, we get:

On the right, there are terms, thus:

And so:

Discarding the negative root, we are left with:

Does this make more sense?