Question

How do you use pascals triangle to expand `(2y-x)^5`?

Answer 1

`32y^5-80x y^4+80x^2 y^3-40x^3 y^2+10x^4 y-x^5`

Explanation:

A picture of Pascal's Triangle is shown below.

Make sure you note how this picture is made. The numbers in each row are the sum of the two numbers most directly above it (or, in the case of the 1's, the one number most directly above them).

The binomial theorem says that, to expand an expression of the form `(a+b)^n`, you should go to the `n+1^{\mbox{st}}` row of the triangle and use the numbers in that row as coefficients of corresponding products `a^{p}b^{n-p}`, and add these products with the corresponding coefficients as `p` ranges from `n` down to `0`.

So, for `n=5`, we go to the sixth row (1, 5, 10, 10, 5, 1) to say:

`(a+b)^{5}=a^{5}+5a^{4}b+10a^[3}b^{2}+10a^{2}b^{3}+5ab^{4}+b^{5}`

Now do the same thing with `a=2x` and `b=-y` to get

`(2x-y)^{5}=(2x)^{5}+5(2x)^{4}(-y)+10(2x)^{3}(-y)^{2}+10(2x)^{2}(-y)^{3}+5(2x)(-y)^{4}+(-y)^{5}`

`=32y^5-80x y^4+80x^2 y^3-40x^3 y^2+10x^4 y-x^5`