Math Portfolio higher level type 1

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.