Since I have already found out the numerators (Step 1), the complete 6th and 7th row are as followed:

6th row : 1 1

7th row : 1 1

Step 3:

Step 3 is about finding a general expression for the numerator. In Figure 1, we see that the relation between the numerator and n is must be a nonlinear equation. So I use a general quadratic approach to find the exact expression for the numerator as a function of row number n.

Numerator(n)=an²+bn+c

As we have 3 unknown variables a, b and c, we need 3 independent equations. To find these equations, we simply put 3 already known numerators with the associated row number. Therefore I chose row number 4 , 5 , 6.

Equation I: 10=16a+4b+c

Equation II: 15=25a+5b+c

Equation III: 21=36a+6b+c

To eliminate c, we subtract equation I from equation II and equation II from Equation III.

IV:Equation II –Equation I => 5=9a+b

V:Equation III – Equation II=> 6=11a+b

Now we subtract equation IV from equation V:

V-IV => 1=2a =>a=0.5

To find b, we put a in equation IV:

5=9x0.5+b

5=4.5+b

b=0.5

To find our last unknown variable c, we put a and b in I:

10=16x0.5+4x0.5+c

10=10+c

c=0

Now after we found all three unknown variables, we put them in our original, general quadratic equation.

Numerator=0.5n²+0.5n

Numerator=

To check, if our expression is correct, we check put in a row number and check if the associated numerator is correct:

Validation test 1:

n=3

Numerator=(3²+3)/2

Validation test 2:

n=6

Numerator=(6²+6)/2

Validation test 3:

n=7

Numerator=(7²+7)/2

Step 4:

The next step is , to find a general expression for the denominator as a function of row number n and element number r.

First of all I’m going to show how the denominator for r=1 (the red circled numbers).

Figure 6 is showing the relation between the denominator and n

To find the general expression for the denominator, we have to use the general, quadratic approach again:

Denominator=an²+bn+c

We need three equations to find all three variables. Therefore we put in the numbers from row 3, 4 and 5 and their associated denominators. Remember that we for now look at the red circled numbers

Quadratic equation for r=1

I:4=9a+3b+c

II:7=16a+4b+c

III:11=25a+5b+c

IV:III-II =>4=9a+b

V:II-I =>3=7a+b

IV-V:1=2a=>a=0.5

a in IV: 4=9x0.5+b=>b=-0.5

a and b in I: 4=9x0.5+3x(-0.5)+c =>c=1

For r=1 the equation for the denominator=0.5n²-0.5n+1

Quadratic equation for r=2

I: 4=9a+3b+c

II: 6=16a+4b+c

III: 9=25a+5b+c

IV: III-II =>3=9a+b

V: II-I =>2=7a+b

IV-V: 1=2a=>a=0.5

a in IV =>3=9x0.5+b=>b=-1.5

a and b in I: 4=9x0.5+3x(-1,5)+c =>c=4

For r=2 the equation for the denominator=0.5n²+(-1,5xn)+4

Quadratic equation for r=3

I:7=16a+4b+c

II:9=25+5b+c

III:12=36a+6b+c

IV: III-II =>3=11a+b

V: II-I =>2=9a+b

IV-V: 1=2a=>a=0.5

a in IV =>3=11x0.5+b=>b=-2.5

a and b in I=> 7=16x0.5+4x(-2,5)=>c=9

For r=3 the equation for the denominator=0.5n²-2.5n+9

As we found the three equations for r=1, 2, 3 we try to find a relationship to r:

r=1 =>Denominator=0.5n²-0.5n+1

r=2 =>Denominator=0.5n²-1.5n+4

r=3 =>Denominator=0.5n²-2.5n+9

The first part of the equation is similar to the equation of the numerator. Now we change the equations, that the first part of the equations looks like the numerator. Therefore we add 0.5n and

-0.5:

r=1 D=0.5n²-0.5n+1-0.5n+0.5=>D=0.5n²+0.5n+1-1n

r=2 D=0.5n²-1.5n+4+0.5n-0.5n=>D=0.5n²+0.5n+4-2n

r=3 D=0.5n²-2.5n+9+0.5n-0.5n=>D=0.5n²+0.5n+9-3n

As we can see now, the last part of the equation always has the form r²-rn. Therefore I can say, that the general expression for the denominator is as followed:

D=Numerator-r(n-r)

D=

Step 5:

Our next step is it, to find the general statement for En(r) and validate it. To find the general statement, I simply need to divide the numerator by the denominator.

En(r)=

En(r)=

In order to test the validity, I used the equation to find the eighth and ninth row.

Row eighth:

Row ninth:

1 1

As I’ve found out the patterns for our set of numbers, I can see, that the equation works. But to make it a 100% sure, I am going to use the equation on elements given in the given set of numbers:

E5(2)==

E4(3)==

Step 6:

In step 6 we are going to discuss n and r and their specific limitations.

Since n is the row number, it starts with 1 and is going up to infinity.

So n=[1; ∞[

The limitations for r are, that r is somewhere between 0 and not more than its specific row number n.

So in line n, r=[0;n]

Step 7:

I will shortly explain, how I arrived at my general statement.

At first, I had a look at the given set of number, where I found patterns between the numerators and the row numbers and also for the denominator and the row numbers.

With that information I came up with the approach using the general quadratic equation. For the numerator it wasn’t too difficult because I only had to solve 3 different equations in order to get the three unknown variables for Numerator (n) =an²+bn+c.

For the denominator, it was a bit trickier. The first step was similar. Using the denominators values to solve Denominator=an²+bn+c. After solving those three equations, I noticed that the first part of the equation looked similar to the general numerator expression. So I tried to make it the same by adding +0,5n-0,5n in order to get 0.5n²+0,5n in each equation. After that was done, I saw that the last part of the equation was always r²-rn.

After finding the formula for the numerator and denominator, finding the general statement for En(r) wasn’t much of a big deal. All I had to do was dividing the numerator by the denominator.