Question
Expand (x-1/2y)^5 by using combination method
Answers
GPT 3.5
To expand (x - 1/2y)^5 using the combination method, we can use the binomial theorem. The binomial theorem states that (a + b)^n = C(n, 0)a^n*b^0 + C(n, 1)a^(n-1)*b^1 + C(n, 2)a^(n-2)*b^2 + ... + C(n, n-1)a^1*b^(n-1) + C(n, n)a^0*b^n, where C(n, k) is the combination formula which calculates the number of ways to choose k items from a set of n items.
In this case, a = x and b = -1/2y. So, the expansion of (x - 1/2y)^5 will be:
C(5, 0)x^5(-1/2y)^0 + C(5, 1)x^4(-1/2y)^1 + C(5, 2)x^3(-1/2y)^2 + C(5, 3)x^2(-1/2y)^3 + C(5, 4)x^1(-1/2y)^4 + C(5, 5)x^0(-1/2y)^5
Now, let's calculate each term:
C(5, 0)x^5(-1/2y)^0 = 1*x^5*(-1/2y)^0 = x^5
C(5, 1)x^4(-1/2y)^1 = 5*x^4*(-1/2y)^1 = -5/2*x^4y
C(5, 2)x^3(-1/2y)^2 = 10*x^3*(-1/2y)^2 = 10/4*x^3y^2 = 5/2*x^3y^2
C(5, 3)x^2(-1/2y)^3 = 10*x^2*(-1/2y)^3 = -10/8*x^2y^3 = -5/4*x^2y^3
C(5, 4)x^1(-1/2y)^4 = 5*x^1*(-1/2y)^4 = 5/16*x^1y^4
C(5, 5)x^0(-1/2y)^5 = 1*x^0*(-1/2y)^5 = -1/32*y^5
Putting it all together, the expansion of (x - 1/2y)^5 is:
x^5 - 5/2*x^4y + 5/2*x^3y^2 - 5/4*x^2y^3 + 5/16*x^1y^4 - 1/32*y^5
In this case, a = x and b = -1/2y. So, the expansion of (x - 1/2y)^5 will be:
C(5, 0)x^5(-1/2y)^0 + C(5, 1)x^4(-1/2y)^1 + C(5, 2)x^3(-1/2y)^2 + C(5, 3)x^2(-1/2y)^3 + C(5, 4)x^1(-1/2y)^4 + C(5, 5)x^0(-1/2y)^5
Now, let's calculate each term:
C(5, 0)x^5(-1/2y)^0 = 1*x^5*(-1/2y)^0 = x^5
C(5, 1)x^4(-1/2y)^1 = 5*x^4*(-1/2y)^1 = -5/2*x^4y
C(5, 2)x^3(-1/2y)^2 = 10*x^3*(-1/2y)^2 = 10/4*x^3y^2 = 5/2*x^3y^2
C(5, 3)x^2(-1/2y)^3 = 10*x^2*(-1/2y)^3 = -10/8*x^2y^3 = -5/4*x^2y^3
C(5, 4)x^1(-1/2y)^4 = 5*x^1*(-1/2y)^4 = 5/16*x^1y^4
C(5, 5)x^0(-1/2y)^5 = 1*x^0*(-1/2y)^5 = -1/32*y^5
Putting it all together, the expansion of (x - 1/2y)^5 is:
x^5 - 5/2*x^4y + 5/2*x^3y^2 - 5/4*x^2y^3 + 5/16*x^1y^4 - 1/32*y^5