A Mathematical Analysis Of Pascal's Fibonacci

Through Pascal’s triangle one can determine the Fibonacci numbers. Pascal’s triangle is a number triangle that arranges its terms in staggered rows, beginning with the very top row (n=0 or the 0^(th) row). It is named after the French mathematician, Blaise Pascal and a variety of other mathematicians and scholars, explored the usages and the possibilities of the triangle. Nowadays, it serves in many mathematical situations like: the binomial and trinomial expansion in order to determine the coefficients, in probability, determining possible combinations and also polynomials.
Image 3. Pascal’s triangle (rows 0-8)

Each value in the triangle is the sum of the two numbers directly above it, as demonstrated in the image below.
Image 4. Explanation of the relationship between the values in Pascal’s triangle (excluding the outer layers of 1s)

Pascal’s triangle and the Fibonacci numbers relate to each other because it is a property of the triangle, that the sum of its diagonals equates to the Fibonacci numbers, as can be seen in the following image.

Image 5. Relationship between Pascal’s triangle and Fibonacci numbers Therefore, I can claim that the sum of the terms in the nth diagonal of Pascal’s triangle, must be equal to the nth Fibonacci number, for all positive integers n>0.
This is more clearly and mathematically depicted in the following modification of the triangle.
Image 6. Modified Pascal’s

Through mathematical induction, Binet’s formula may also be proven.
Mathematical induction is a method used to prove a statement/theory that is applicable for all natural numbers.
Let n=5
F_5= ((〖(1+√5)〗^5-(〖1-√5)〗^5)/(2^5 √5))
Therefore, F_5= 5.04 ≈5
Which is correct. Let F_n represent Fibonacci number n.
Where, F_0=1 and F_1=1.
F_n=1/√5 〖[((1+√5)/2)〗^n-((1-√5)/2)^n]
F_n= ((〖(1+√5)〗^n-(〖1-√5)〗^n)/(2^n √5))
Let n= 1
F_1= ((〖(1+√5)〗^1-(〖1-√5)〗^1)/(2^1 √5))

Therefore, F_1=1
Which is correct.
Let n= 3
F_3=((〖(1+√5)〗^3-(〖1-√5)〗^3)/(2^3 √5))
Therefore, F_3=2
Once again, this is correct.

So to conclude, both methods of finding the n^(th)Fibonacci term give the same results but have different efficiencies. Binet’s formula is definitely the more effective of the two as it is less time consuming and much clearer than using Pascal’s triangle.
The Fibonacci sequence, though it may seem complex and useless, has proven to be extremely useful and relevant in the day-to-day world. It incorporates so many other mathematical concepts, like the golden ratio, geometrical sequences, quadratic functions, etc. which makes it extremely useful in the mathematical