Answer:
A) (3x + y)^4:
The Binomial Theorem states that for any real numbers a and b, and any positive integer n, the expansion of (a + b)^n is given by the sum of n terms of the form:
C(n,k)a^(n-k)b^k
where C(n,k) is the binomial coefficient (n choose k).
So, expanding (3x + y)^4:
(3x + y)^4 = C(4,0)(3x)^4(y)^0 + C(4,1)(3x)^3(y)^1 + C(4,2)(3x)^2(y)^2 + C(4,3)(3x)^1(y)^3 + C(4,4)(3x)^0(y)^4
Using the binomial coefficients:
C(4,0) = 1, C(4,1) = 4, C(4,2) = 6, C(4,3) = 4, C(4,4) = 1
Expanding the terms:
(3x + y)^4 = 1(3x)^4(y)^0 + 4(3x)^3(y) + 6(3x)^2(y)^2 + 4(3x)(y)^3 + 1(y)^4
= 1(81x^4) + 4(27x^3y) + 6(9x^2y^2) + 4(3xy^3) + 1(y^4)
= 81x^4 + 108x^3y + 54x^2y^2 + 12xy^3 + y^4
B)
The Binomial Theorem states that for any real numbers a and b and any positive integer n, the expansion of (a + b)^n is given by:
(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) represents the binomial coefficient, which can be calculated as n! / (k! (n - k)!).
Applying this to (x - 2y)^6, we get:
(x - 2y)^6 = C(6, 0)x^6 * (-2y)^0 + C(6, 1)x^5 * (-2y)^1 + C(6, 2)x^4 * (-2y)^2 + ... + C(6, 5)x^1 * (-2y)^5 + C(6, 6)x^0 * (-2y)^6
Expanding each term:
C(6, 0) = 6! / (0! 6!) = 1
C(6, 1) = 6! / (1! 5!) = 6
C(6, 2) = 6! / (2! 4!) = 15
C(6, 3) = 6! / (3! 3!) = 20
C(6, 4) = 6! / (4! 2!) = 15
C(6, 5) = 6! / (5! 1!) = 6
C(6, 6) = 6! / (6! 0!) = 1
Now we can substitute each value into the equation:
(x - 2y)^6 = 1x^6 - 12x^5 * 2y + 90x^4 * (-2y)^2 - 360x^3 * (-2y)^3 + 720x^2 * (-2y)^4 - 720x * (-2y)^5 + 1 * (-2y)^6
Answer:
(x - 2y)^6 = x^6 - 12x^5 * 2y + 90x^4 * (-2y)^2 - 360x^3 * (-2y)^3 + 720x^2 * (-2y)^4 - 720x * (-2y)^5 + (-2y)^6.