Let
Now by plugging the following values in the expression we check if it is divisible by 5.
which is divisible by 5
Which is divisible by 5.
Therefore is divisible by 5.
Using mathematical induction I'm going to prove whether is always divisible by the that was divisible by. The values of that was divisible by were 2, 3 and 5. Therefore using mathematical induction I'm going to check if is always divisible for these values.
The First case is
Now by using mathematical induction we want to prove that is divisible by 2.
First we let n=1 and see if it is divisible by 2
Therefore divisible by 2,
Now we assume that is true so is true.
However, we need to prove that is true.
And .
Now let's subtract from.
The expression is now
Let's solve the brackets, the expression becomes in this form:
By collecting terms and simplifying the
Expression it will become:
Which is indeed divisible by 2
The second case is
Now by using mathematical induction we want to prove that is divisible by 3.
First we let n=1 and see if it is divisible by 3
Therefore divisible by 3.
Now we assume that is true so is true.
However, we need to prove that is true.
And .
Now let's subtract from.
The expression is now
Let's solve the brackets so the expression becomes in this form:
By collecting terms and simplifying the
Expression it will become:
And by taking as common factor the expression is now:
Therefore it is divisible by 3
Since is not divisible by 4 it is ignored and we don't have to prove by induction.
The 3rd case is when
Now by using mathematical induction we want to prove that is divisible by 5.
First we let n=1 and see if it is divisible by 5
Therefore divisible by 5.
Now we assume that is true so is true.
However, we need to prove that is true.
And .
Now let's subtract from.
The expression is now
Let's solve the brackets, so the expression becomes in this form:
By collecting terms and simplifying the
Expression it will become:
And by taking as common factor the expression is now:
Therefore it is divisible by 5.
Now let's explore more cases for and if is divisible bywe'll prove it by induction. So we will factorize the expression for
Let's look at when
Now let's plug the value of in the expression.
Now let's keep factorizing the expression more
I don't find any clear evidence in the expression to show if the expression is divisible by 6 or not however, let's take a few example to check.
Let
Now by plugging the following values in the expression we check if it is divisible by 6.
which is not divisible by 6
Which is not divisible by 6.
Therefore is not divisible by 6 which is not prime.
Let's look at when
Now let's plug the value of in the expression.
Now let's keep factorizing the expression more
I don't find any clear evidence in the expression to show if the expression is divisible by 7 or not however, let's take a few example to check.
Let
Now by plugging the following values in the expression we check if it is divisible by 7.
which is divisible by 7
Which is divisible by 7.
Therefore is divisible by 7.
Now by using mathematical induction we want to prove that is divisible by 7.
First we let n=1 and see if it is divisible by 7
Therefore divisible by 7.
Now we assume that is true so is true.
However, we need to prove that is true.
And .
Now let's subtract from.
The expression is now
Let's solve the brackets so the expression becomes in this form:
By collecting terms and simplifying the expression it will become:
And by taking as common factor the expression is now:
Therefore it is divisible by 7.
Let's look at when
Now let's plug the value of in the expression.
Now let's keep factorizing the expression more
I don't find any clear evidence in the expression to show if the expression is divisible by 11 or not however, let's take a few example to check.
Let
Now by plugging the following values in the expression we check if it is divisible by 11.
Which is divisible by 11
Which is divisible by 11.
Therefore is divisible by 11.
Now by using mathematical induction we want to prove that is divisible by 11.
First we let n=1 and see if it is divisible by 11
Therefore divisible by 11.
Now we assume that is true so is true.
However, we need to prove that is true.
And.
Now let's subtract from.
The expression is now
Let's solve the brackets so the expression becomes in this form:
By collecting terms and simplifying the expression it will become:
And by taking as common factor the expression is now:
Therefore it is divisible by 11.
After looking at several cases I came up with a conjecture that if in the expression is a prime number then is divisible by .
Now let's look at Pascal's triangle to find a pattern and explain how the entries are obtained.
Pascal's rules states that using the binomial coefficient which is where is the number of rows and is the position in the row. Starting with =0. For example in order to obtain the second entry in the 2nd row we plug the following in this equation which is equal to 2. In the Pascal triangle each row starts and ends with 1 and there is infinite number or rows. And the entries increases by 1 as you increase the number of rows. For example, the first row has 0 entries and the second row has 1 entries and the third has 2 entries and the forth has 3 entries and so on. So if you want to generate the 3rd row for example you know that it has 2 entries and it start with 1 and ends with 1 so you need to find the first and 2nd entry. You can find them using Pascal's rule. The first entry is 3 using the Pascal's rule where you plug the number of row and the entry's number in the equation. So the first entry in the 3rd row is which is equal to 3 and the second entry is found by plugging the number of rows and the position in the row in the equation and it will turn out to be: which is also equal to 3. Therefore the first entry will be 3 and the second entry is 3. Thus, the row will look like this 1 3 3 1.
Using GDC I can apply Pascal's rule using the where is the number of rows and is the position in the row. And so I can generate the first 15 rows using GDC. For example, let's say that I want to generate the 3rd row, I know that it has 2 entries and that it starts with 1 and ends with 1.Therefore in order to find the first entry I click on 3 on the calculator as it is the number of rows and then I click on the math button and go to the PRB tab scrolling down to the third option and then I click the on the number of entry I want to find. In this case 1 and so it will look like this on the calculator screen 31. Afterwards I click enter and it will give me 3 which is the first entry in the 3rd row. I will repeat this process for the first 15 rows and I can find them using this technology.
Here are the first 15 rows of the Pascal's triangle:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
I found a relationship between Pascal's triangle and the expression. The coefficients of in the expression are the entries in the row of Pascal's triangle and is the number of rows. For example if you want to find the entries of the 5th row you plug 5 in the expression instead of so the expression is now by simplifying the expression it will look like this: and 5 10 10 5 are the entries in the fifth row and since each row starts and ends with 1 you can find the row and its entries by this relation. So the fifth row of Pascal's triangle will look like this 1 5 10 10 5 1. In addition, there's another relationship between Pascal's triangle and , if the number of rows is a prime number the entries in the row will be divisible by . For example, let's look at the 3rd row of Pascal's triangle. The entries in the 3rd row 1 3 3 1 are divisible by 3 since 3 is a prime number. The same is with the expression if is a prime number then the expressionis divisible by. Therefore, from this relationship we can conclude that is a multiple of if is a prime number.
After finding this relationship I came up with a new conjecture. If in the expression is a prime number then is divisible by . Also, if is a prime number in Pascal's triangle where is the number of rows then the entries in the row are divisible by and it is a multiple of.
However, if we look at the converse of this statement which is if expression divides by then is a prime number. This statement is not true since this expression can divide by some non-prime numbers for some values of.
For example let's take 4, although 4 is not prime yet if you plug 5 in the expression instead of it will divide by 4.
which is divisible by 4.
Therefore, the converse of my conjecture is not true and doesn't hold.