The Trinomial Theorem and Pascal’s Tetrahedron

Like most situations, there are two ways in which you can look at things. Consider this situation with the idea of a coin. On one side, you can use the trinomial expansion theorem to determine the coefficients of terms within Pascal’s Tetrahedron. On the other hand, you can use the already existent Pascal’s Triangle to derive the coefficients in Pascal’s Tetrahedron and from then conduct the trinomial expansion with great ease. Think of it this way: you can use algebra to help you with geometry, but you can also use geometry to help you with algebra. Anyway, however, you look at it, you come to the same conclusion; Blaise Pascal’s level of ingenuity continues to rise with each passing day, as the patterns found in Pascal’s tetrahedron are just as fascinating as those found in the triangle.

The definition of Pascal’s tetrahedron (or triangular pyramid) is as follows: Pascal’s pyramid is a three-dimensional arrangement of the trinomial numbers, which are the coefficients of the trinomial expansion and the trinomial distribution”. The tetrahedron is a correspondent of the two-dimensional Pascal’s Triangle. The problem, with the tetrahedron is that it can be extremely hard to visualize because of its three dimensional nature. Another issue arises when you attempt to derive the trinomial coefficients that would fit in the tetrahedron. This is because of the inability to visualize the tetrahedron. To begin with, let us derive the first four layers (zero to three) of Pascal’s Tetrahedron, using algebra and expansion. This is done by firstly expanding each of the trinomials linearly, and then organizing them in a theoretically understandable manner, creating layers of Pascal’s Tetrahedron.

First Four Layers (Zero to Three) of Pascal's Tetrahedron

From the illustration shown above, you may notice the following: the last rows of each layer are synonymous with their respective rows in Pascal’s Triangle. For instance, as the illustration shows below, the last row of the fourth layer of Pascal’s Tetrahedron reads as: 1 3 3 1, which is the same as the fourth row in Pascal’s triangle.  In addition, you will also realize that the diagonals adjoining the vertices also read 1 3 3 1, which corresponds to the fourth row in Pascal’s triangle. Moreover, you will notice that each outline of the nth layer in Pascal’s Triangle is a binomial expansion of degree n. As we can see from the example shown below of the fourth layer of Pascal’s Tetrahedron, the left outline is a binomial expansion of (a+b)3, while the right outline is a binomial expansion of (a+c)3 and the bottom outline is a binomial expansion of (b+c)3. Each of them is highlighted in yellow for identification purposes. Furthermore, you may notice that the terms with the highest powers are situated at the vertices, because the coefficients of those terms are “1”. This is analogous to the binomial theorem which utilizes Pascal’s Triangle.

Pascal's Triangle within Pascal's Tetrahedron

Another interesting pattern present in Pascal’s Tetrahedron that can be related to Pascal’s Triangle, is the sum of the terms within each layer. In Pascal’s Triangle, which was two-dimensional, the expression used to determine the sum of the row was (1+1)n, each “1” pertaining to the number of dimensions, this was then simplified down to 2n. Similarly, as we have added another dimension, the sum of the terms present in the nth layer of Pascal’s Tetrahedron is attained using the expression (1+1+1)n, which is simplified down to 3n.  Moreover, the number of terms in the nth layer is equal to (n+1)(n+2)/2.

Verification (1): The third layer of Pascal’s Tetrahedron has the terms: (1 3 3 3 6 3 1 3 3 1), which adds up to a total of 27. In this case, n = 3, because it is the third layer. Therefore, the expression which represents the sum of the terms of the third layer is 33, which results in 27. This proves that the expression 3n can be used to determine the sum of the terms present in the nth layer of Pascal’s Tetrahedron.

Verification (2): Q) How many terms are present in the 5th layer of Pascal’s Tetrahedron.  A) n = 5, ((5+1) x (5+2))/2 = (6)(7)/2 = 42/2 = 21. By counting the number of terms in the 5th layer of Pascal’s Tetrahedron, you will realize that there are exactly 21 terms, and that this formula works.

In order to visualize Pascal’s Tetrahedron, as well as try and come up with the coefficients for the trinomial theorem,  try and picture each the layers sitting on top of each other. Don’t feel like you are the only one struggling to picture this, I had issues too. You may wonder, don’t we just add the two numbers above, and come up with the corresponding coefficient. Well, unlike Pascal’s Triangle, this is slightly more complicated. In order to attain the next entry in the nth layer of Pascal’s Tetrahedron, you must sum the entries in the nth-1 layer that are touching it. Basically, the goal is to form an equilateral triangle, and sum the three numbers to get the next entry. In simpler terms, the sum of three adjacent numbers in the previous layers gives it a number in the consequent layer. However, if the third adjacent number falls outside the triangular layer, that number is represented as a zero (see blue and orange highlights). In the illustration shown below, the outcome, which is the term in the sixth layer, is highlighted in one color and the terms required to reaching that term are highlighted in the same color in the fifth layer.

Deriving Terms

In class, we went over how to use Pascal’s Triangle’s coefficients and relate them to the binomial theorem. The formula, that is used determine the trinomial coefficients and exponent values in Pascal’s Tetrahedron, is quite similar to the one used to determine binomial coefficients and exponent values in Pascal’s Triangle. For binomial expansions where (am + bk) n, m + k must equal n, at all times, while the value of the coefficient was either taken directly from Pascal’s Triangle or calculated using the formula shown below, where n was the value of the degree (which corresponded to the row number) and m was the term number. For trinomial expansions, in the form of (am+ bk + cl)n, for all terms, m+k+l = n, while the value of the coefficient can be derived using Pascal’s Tetrahedron or using the formula shown below, which is an extension of the formula used to determine binomial coefficients. In more mathematical terms, the value of the coefficient of Pascal’s Tetrahedron located at (n-m, m-k, k) which is the coefficient of the term an-mbm-kck can be obtained using the trinomial coefficient formula shown below. The formula for the trinomial theorem is also shown below, and is partially, if not entirely composed of the formula to determine trinomial coefficients.

Coefficients Formula (Binomial and Trinomial)

Trinomial Theorem

Putting Pascal’s Tetrahedron and The Trinomial Theorem To Work:

Question: Expand (a+b+c)4
Answer: There are two ways to do this. A) Derive the coefficients using Pascal’s Tetrahedron or B) Use the Trinomial Coefficients Formula to derive the coefficients.
Using Method B – Steps (See Solution Below):
1. Ensure that the highest powers of a, b, c each lie on the vertices of the triangle
2. Ensure that along the edges of either side, you are decreasing the value of (n-m) and increasing the value of either (m-k) or k.
3. For the base, ensure that the values of (m-k) decreases, while the value of k increases as you shift from left to right.
4. Fill in the exponents of a, b and c, in a similar fashion.
5. Compute and sum each of the terms to obtain the expanded form of (a+b+c)4.

Solution (a+b+c)^4

Fun For You
a) Expand (a+b+c)7 using Method B (stated above)
b) Derive a formula for i) the number of terms for a multinomial element ii) the sum of the terms for a multinomial element iii) the coefficient formula for a multinomial expansion.



8 thoughts on “The Trinomial Theorem and Pascal’s Tetrahedron

  1. wow. dis blg ws rlly helpful 2 me. tried srchin d net 4 smth lyk this, cldnt find anyth
    very clr analys, i am abl to undstnd now

  2. I can see you happen to be an expert at your field! I am launching a site soon, and your facts is going to be very useful for me.. Thanks for all your aid and wishing you all of the success.

  3. I actually worked this out and played with it in high school, but I didn’t stop with 3 dimensions. The same Pascal extensions apply to any number of polynomials, generating a triangle structure with as many physical dimensions as the number of terms. I spent some wasted time constructing a 4 dimensional hyperhedron, most easily represented as a stack of tetrahedrons, one for each layer. One other quality I found while playing with this was that the sum of the coefficients of any layer is equal to the number of the terms raised to the power of number of the layer. Thus, in a simple Pascal’s Triangle, (a+b)^3 = 1 3 3 1 which totals 8 = 2^3. or (a+b)^5 = 1 5 10 10 5 1 = 32 = 2^5, but the same still holds true in higher dimensions, so for example (a+b+c+d+e)^4 works out to 5^4 or 625. Never did do anything with it, but the adding-adjacent system for working out the next lower level got very interesting in the 4th & 5th dimensions.

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