`(x + 2y)^4` Use the Binomial Theorem to expand and simplify the expression.

You need to use the binomial formula, such that:

`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k`

You need to replace x for x, 2y for y and 4 for n, such that:

`(x+2y)^4 = 4C0 x^4 + 4C1 x^3*(2y)^1 + 4C2 x^2*(2y)^2 +  4C3 x*(2y)^3 +  4C4 (2y)^4 `

By definition, `nC0 = nCn = 1` , hence `4C0 = 4C4 = 1` .

By definition `nC1 = nC(n-1) = n` , hence `4C1 = 4C3 = 4.`

By definition `nC2 = nC(n-2) = (n(n-1))/2` ,...

You need to use the binomial formula, such that:

`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k`

You need to replace x for x, 2y for y and 4 for n, such that:

`(x+2y)^4 = 4C0 x^4 + 4C1 x^3*(2y)^1 + 4C2 x^2*(2y)^2 +  4C3 x*(2y)^3 +  4C4 (2y)^4 `

By definition, `nC0 = nCn = 1` , hence `4C0 = 4C4 = 1` .

By definition `nC1 = nC(n-1) = n` , hence `4C1 = 4C3 = 4.`

By definition `nC2 = nC(n-2) = (n(n-1))/2` , hence `4C2 = (4(4-1))/2 = 6`

`(x+2y)^4 = x^4 + 4x^3*(2y)^1 + 6x^2*4y^2 +  4x*8y^3 + 16y^4 `

`(x+2y)^4 = x^4 + 8x^3*y + 24x^2*y^2 +  32x*y^3 + 16y^4 `

Hence, expanding the complex number using binomial theorem yields the simplified result `(x+2y)^4 = x^4 + 8x^3*y + 24x^2*y^2 +  32x*y^3 + 16y^4.`