Use the binomial theorem to expand (2x-3y)^5
showing work is appreciated
(2x)^5 + 5C1 (2x)^4 (-3y)^1 + 5C2 (2x)^3 (-3y)^2 + 5C3 (2x)^2 (-3y)^3 +5C4 (2x)^1 (-3y)^4 + (-3y)^5
Expand (2x+3y)`5
I want u to expand for me this question (2x+3y)5 using the Pascal's triangle
Student
To expand the expression (2x - 3y)^5 using the binomial theorem, we need to find the coefficients of each term. The binomial theorem states that (a + b)^n = ∑(k=0 to n) [nCk * a^(n-k) * b^k], where nCk represents the combination of "n choose k" and is given by n!/(k!(n-k)!).
In this case, our expression is (2x - 3y)^5, so a = 2x and b = -3y, and n = 5. Let's calculate each term step by step:
Term 1 (k = 0):
(nCk * a^(n-k) * b^k) = (5C0) * (2x)^(5-0) * (-3y)^0 = 1 * (2x)^5 * 1
Term 2 (k = 1):
(nCk * a^(n-k) * b^k) = (5C1) * (2x)^(5-1) * (-3y)^1 = 5 * (2x)^4 * (-3y)^1
Term 3 (k = 2):
(nCk * a^(n-k) * b^k) = (5C2) * (2x)^(5-2) * (-3y)^2 = 10 * (2x)^3 * (-3y)^2
Term 4 (k = 3):
(nCk * a^(n-k) * b^k) = (5C3) * (2x)^(5-3) * (-3y)^3 = 10 * (2x)^2 * (-3y)^3
Term 5 (k = 4):
(nCk * a^(n-k) * b^k) = (5C4) * (2x)^(5-4) * (-3y)^4 = 5 * (2x)^1 * (-3y)^4
Term 6 (k = 5):
(nCk * a^(n-k) * b^k) = (5C5) * (2x)^(5-5) * (-3y)^5 = 1 * (2x)^0 * (-3y)^5
Now, let's simplify each term:
Term 1: 1 * (2x)^5 * 1 = 2^5 * x^5 = 32x^5
Term 2: 5 * (2x)^4 * (-3y)^1 = 5 * 2^4 * x^4 * (-3)^1 * y^1 = -240x^4y
Term 3: 10 * (2x)^3 * (-3y)^2 = 10 * 2^3 * x^3 * (-3)^2 * y^2 = 720x^3y^2
Term 4: 10 * (2x)^2 * (-3y)^3 = 10 * 2^2 * x^2 * (-3)^3 * y^3 = -2160x^2y^3
Term 5: 5 * (2x)^1 * (-3y)^4 = 5 * 2^1 * x^1 * (-3)^4 * y^4 = 810x^1y^4
Term 6: 1 * (2x)^0 * (-3y)^5 = 1 * 2^0 * (-3)^5 * y^5 = -243y^5
Putting it all together, the expanded form of (2x-3y)^5 is:
32x^5 - 240x^4y + 720x^3y^2 - 2160x^2y^3 + 810xy^4 - 243y^5