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

As with many proofs related to Pascal’s triangle, the connection to the binomial expansion can be proven inductively. However, in this article, we are not going to worry too much about what proof by induction is, nor use the notation or terminology that often accompanies it. I’m just going to explain the logic 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 generalized to cover any power of (a+b ).

In this case, consider (a+b)^6. It can be done from the expansion of (a+b)^5 all multiplied by another parenthesis 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)

It seems complicated and unrevealing. However, we can now begin to think about how this all relates to Pascal’s triangle. For example, let’s say we are trying to do 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 either by **a** or **b**. That leaves us with two ways to achieve our goal of getting a^2b^4. We could choose a^2b^3 and multiply it by **b** from outside the parenthesis, or we could start with ab^4 and multiply it by the **a** from the outside of the support.

Since 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 how many lots of a^2b^3 of the parenthesis there are, plus how many lots of ab^4 there are has. 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 leads. 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 statement we are trying to prove, we also state that you add one to the power of b to see how many numbers you need to count to get the desired coefficient. Thus, 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 on the 6th line. If you think about it, these three numbers are positioned in such a way that they form a small triangle of numbers in Pascal’s triangle.

Applying these arguments to any situation, we can see that Pascal’s triangle rules will hold for any term of any power of (a + b), because we can always divide our term into coefficients of two separate terms positioned adjacently directly and directly above our initial term. The only case where 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 it obviously works anyway.

Had finished! It wasn’t an easy proof, so congratulations on getting there! Even if you haven’t got it all figured out, I hope this has given you some insight into how the seemingly unrelated topic of expansion and binomial expansion is actually so closely related to Pascal’s triangle.

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