ArtsAutosBooksBusinessEducationEntertainmentFamilyFashionFoodGamesGenderHealthHolidaysHomeHubPagesPersonal FinancePetsPoliticsReligionSportsTechnologyTravel

Math Series Part VII: Pascal's Triangle

Updated on November 30, 2016
Michael Ttappous profile image

Michael has been an online freelancer and writer for many years and loves discovering and sharing about new experiences and opportunities.

Pascal’s triangle presents us with a very interesting case study on numbers. Indeed, the triangle itself, constructed of numbers ordered in rows such that they form the binomial coefficients, contains within it many other demonstrable sequences and patterns. Alongside that of symmetry and the obvious addition that occurs, there are also less observable qualities that are truly magnificent. If we look at this figure, for example, we can see, by observing the triangle diagonally, that there is a sequence of ones, and then a sequence of positive integers, and then we get to a sequence of triangle numbers. Triangle numbers themselves count the amount of homogenous objects needed in order to form an equilateral triangle. If we consider one object, T1, to be equilateral, then for T2 we need to add two objects to the base. For T3, we add another 3; for T4 another four to the base. We therefore get a sequence of triangle numbers of 1, 3, 6, 10, 15, 21, 28, 36, 45, etc. Returning back to the picture above, we can see that these appear clearly in the third diagonal of Pascal’s triangle.

The reason they do so is because in order to go from Tn to Tn + 1, we need to add n + 1 number of objects to keep the triangle equilateral. As the red circle above illustrates, the first blue square represents T1. The second blue square represents T2. If we look closely, we will see that T2 - T1 is equal to the yellow square next to the first blue square. Indeed, this is a property of the Pascal triangle itself. And it is because this property is so consistent that it holds true that, for any value Tn, the result is Tn – 1 + the value to the left of Tn – 1. It is due to this property that we find this continuous line of triangle numbers in Pascal’s triangle.

In much the same way as we observe the triangle numbers, we can also find the Fibonacci sequence in Pascal’s triangle by looking at a slightly steeper diagonal. The picture here illustrates how, by slicing through the triangle and adding up the sum of the digits in the line, we get a sequence of the Fibonacci numbers, 1, 1, 2, 3, 5, 8, 13, 21, … If we take a moment to reexamine the Fibonacci sequence, we will find that F0 = F1 = 1 and Fn = Fn−1 + Fn−2 for n ≥ 2. What we also observe about Pascal’s triangle is that there are n rows, with n starting at 0, and k number of objects to choose from that row. This leads us to the general equation of (n k) = n!/k!(n-k)!, where there is precisely this many ways to choose k objects (regardless of the ordering) from n given objects, with 0 ≤ k ≤ n and n ≥ 0.

If we return to and look closely at the picture above, we will see that the sum of the numbers on the third diagonal is the sum of the numbers on the previous two diagonals; i.e. 2 = 1 +1. This is also true for the fourth diagonal: 3 = 2 + 1. Therefore, if we let D(i) represent the sum of the numbers on the Diagonal that represents a Fibonacci number, then we can see that D(i) = D(i - 1) + D(i - 2). And since D(i) = D(i - 1) + D(i - 2) is the exact definition of the Fibonacci numbers, i.e. Fn = Fn−1 + Fn−2, this is proof that the sums of the diagonals in Pascal's triangle represent the Fibonacci numbers, and proof of how a simple triangle can represent some of the world's most intricate mathematical complexities.

Photo Credits:


    0 of 8192 characters used
    Post Comment

    No comments yet.