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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 link to the binomial expansion can be proven inductively. However, in this article, we will not worry too much about what proof by induction is, nor will we use any notation or terminology that often goes with it. I’ll just explain the reason for the expansion of (a+b)^6 being the sixth 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). ).

So, let’s think about (a+b)^6. It can be done 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 seems complicated and not very revealing. However, we can now begin to think about how this all ties into Pascal’s triangle. As an example, let’s say we’re trying to do a^2b^4. As shown in the above expression, 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 for anyone **a** or **b**. This 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 **a** from outside 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 the number of lots of a^2b^3 there are from the bracket, 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’ll see that we have 10a^3b^2 and 5ab^4, so you’ll end up with 15 batches of ‘a^2b^4.

You may now be able to see where this is leading. We only have 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 fourth number along, and ab^4 by the fifth number. Finally, we can say that 15a^2b^4 will be the 5th number in the 6th row. If you think about it, these three numbers are positioned so that they will form a small triangle of numbers in Pascal’s triangle.

Applying these arguments to any situation, we can see that the rules of Pascal’s triangle will apply for any term to any power of (a+b), since we can always divide our term into the coefficients of two separate adjacent terms directly and directly. above our initial mandate. The only time our “split terms” 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.

They were done! This was not an easy test, so congratulations for making it this far! Even if you didn’t fully understand it, hopefully it gave you some insight into how the seemingly unrelated topic of binomial expansion and expansion is so closely tied to Pascal’s triangle.

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