The number of logs is given by:
Can you solve for ?
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.
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?
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: .
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:
Adding the two equations, we find:
On the right, there are terms, thus:
And so:
Discarding the negative root, we are left with:
Does this make more sense?