Proof of the Link Between Pascal’s Triangle and the Binomial Expansion

A famous use of Pascal's triangle is in the Binomial Expansion, the multiplication and simplification of brackets. Less well known, however, is the reason why this intriguing link between these two seemingly unrelated areas of mathematics actually exists. As the binomial expansion deals with multiplying out brackets algebraically, you might think that a proof of its link with Pascal's triangle would also involve a lot a complicated algebra. In this article, however, I have attempted to give an informal proof of this link using as little algebra as possible.

As with many proofs related with Pascal's triangle, the link with the Binomial expansion can be proved inductively. However, in this article, we're not going to worry too much about what proof by induction is, or use any of the notation or terminology that often comes with it. I am just going to explain the rationale behind the expansion of (a+b)^6 being the 6th row of Pascal's triangle, and from there, it should be clear how my arguments can be generalised to cover any power of (a+b).

Right then, let's think about (a+b)^6. It can be made from the expansion of (a+b)^5 all multiplied by another bracket of (a+b), as shown below:

(a+b)^6 = (a+b)(a+b)(a+b)(a+b)(a+b)(a+b) = (a+b)^5(a+b)

= (a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5)(a+b)

= a(a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5) + b(a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5)

This is looking complicated and not very revealing. However, we can now start thinking about how all this links in with Pascal's triangle. As an example, say we are trying to make a^2b^4. As shown in the expression above, to get the terms of (a+b)^6, we can multiply a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 by either a or b. This leaves us with two ways of achieving our goal of getting a^2b^4. We could choose a^2b^3 and multiply it by b from outside the bracket, or we could start with ab^4 and multiply it the a from outside the bracket.

As these are the only two ways, the coefficient of a^2b^4 is the sum of the coefficients of ab^4 and a^2b^3. Basically, the number of lots of a^2b^4 I end up with is the number of lots of a^2b^3 from the bracket there are, plus the number of lots of ab^4 there are. If you look at a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5, you will see that we have 10a^3b^2 and 5ab^4, so we will end up with 15 lots of a^2b^4.

You may now be able to see where this is leading. We just need to think about where these powers are supposed to be represented in Pascal's triangle. 10a^2b^3 and 5ab^4 are both from the expansion of (a+b)^5, so they will be in the 5th row of Pascal's triangle. In the claim we are trying to prove we also state that you add one to the power of b to see how many numbers in you have to count to get your desired coefficient. So the coefficient of a^2b^3 will be represented by the 4th number along, and ab^4 by the 5th number along. Finally, we can say that 15a^2b^4 will be the 5th number in on the 6th row. If you think about it, these three numbers are positioned such that they will form a little triangle of numbers in Pascal's triangle.

Applying these arguments to any situation, we can see that the rules of Pascal's triangle will hold for any term from any power of (a+b), as we can always split our term into the coefficients of two separate terms positioned adjacently directly and directly above our initial term. The only time when our "term-splitting" tactic won't work is for (a+b)^1, but (a+b)^1 = 1a + 1b, and 1,1 is the first row of Pascal's triangle, so this obviously works anyway.

We're done! This was not an easy proof, so congratulations for getting this far! Even if you did not fully understand all of it, hopefully it has given you some insight into how the seemingly unrelated topic of expanding and the binomial expansion is in fact so closely linked to Pascal's triangle.

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