Why and how does the binomial theorem work?
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Ok, let’s start with (x + y)^(2) which is x^(2) + 2xy + y^(2) whose coefficients, 1 2 1, correspond to the third row of Pascal’s triangle.
We’re going to multiply another (x + y), effectively making it (x + y)^(3). This involves multiplying our existing expression, x^(2) + 2xy + y^(2) by (x + y) which is the same taking two copies of that expression, multiplying one by x, the other by y, and adding them all together.
Before adding them together we now have: x^(3) + 2x^(2)y + xy and x^(2)y + 2xy^(2) + y^(3). We can see that multiplying through by x and y individually doesn’t change the coefficients. For both it’s still 1 2 1. But when we add the expressions together, the 1’s at either end (corresponding to the x^(3) and y^(3)) aren’t added to anything and stay as 1’s. The other’s add together giving us 2 + 1 = 3 and 1 + 2 = 3 for a final expression of x^(3) + 3x^(2)y + 3xy^(2) + y^(3) whose coefficients are 1 3 3 1.
And this pattern will hold each time you multiply in another (x + y), meaning the coefficients will follow the pattern shown in Pascal’s triangle.
Imagine you have a job laying tiles. You have laid down a square of 9 tiles, 3 x 3. Your boss comes up to you and tells you to turn that into a 25 tile square, 5 x 5. What do you do?
You add 2 layers of 3 tiles to the top and 2 more to the side. Then you fill in that 2 x 2 corner that did not get filled in by the tiles you already added.
Mathematically that works out by taking 9, 3² or the original square you made, adding 3 x the number of added rows twice, and the number of rows added squared.
If you have any binomial to an integer power (1, 2, 3, etc) it can be written in the form (x + y)^(n).
The resulting expanded polynomial will be the following form:
(x^(n)*y^(0)) + (x^(n-1)*y^(1)) + (x^(n-2)*y^(2)) + … + (x^(0)*y^(n))
Each term will have a coefficient from the n^th row of Pascal’s triangle, which is (please excuse the formatting of the triangle):
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
and so on. So, for the polynomial (x+2y)^4 , we’d get the following:
1([x]^(4)*[2y]^(0)) + 4 ([x]^(3)*[2y]^(1)) + 6([x]^(2)*[2y]^(2)) + 4([x]^(1)*[2y]^(3)) + 1([x]^(0)*[2y]^(4))
which after simplifying becomes
X^(4) + 8x^(3)y + 24x^(2)y^2 + 32xy^3 + 16y^4
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