# Binominal theorem

1. May 19, 2007

### Styx

Use the binomial theorem to expand each of the following. Simpify your answers

(1-y^2)^5

Let a = 1 and b = -(y^2)
Then using binomial theorem, you have:

(a+b)^5 = C(5,0)a^5 + C(5,1)a^4 b + C(5,2)a^3 b^2 + C(5,3)a^2 b^3

+ C(5,4)a b^4 + C(5,5)b^5

Substitute a = 1 and b = -(y^2)

(1-y^2)^5 = 1(1)^5 + 5(1)^4 (-(y^2)) + 10(1)^3 (-(y^2))^2

+ 10(1)^2 (-(y^2))^3 + 5(1)(-(y^2))^4 + 1(-(y^2))^5

= 1 + 5(-(y^2)) + 10y^4 + 10(-(y^6)) + 5y^8 + (-(y^10))

= 1 -5y^2 + 10y^4 -10y^6 -5y^8 -y^10

Does that look right? This is my first go at using binomial theorem...

Last edited: May 19, 2007
2. May 19, 2007

### cepheid

Staff Emeritus
It looks good to me. Your binomial coefficients (the blah "choose" blah factors) make sense if you interpret them as the number of ways of choosing the number of b's out of 5 in each term. You could also formulate it by talking about the number of ways of choosing the number of a's out of 5 in each term. You'd get the same answer, because the binomial coefficients are always symmetric (for example, 5 choose 0 equals 5 choose 5, because the number of ways of choosing all of the elements in a set is the same as the number of ways of choosing none of them: 1. Another example: 5 choose 4 equals 5 choose 1 because the number of ways of choosing 4 elements out of 5 is the same as the number of ways of excluding one of them: 5). This explains why the binomial coefficients build up Pascal's Triangle (all of the rows in the triangle are symmetric and bounded by 1's). I hope this offers further insight.

3. May 19, 2007

### Styx

thanks cepheid