Expand Using Binomial Theorem: (1-y^2)^5

[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...

 

[cepheid]

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.

 

[Styx]

thanks cepheid