- #1
Styx
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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...
(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...
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