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.`
No comments:
Post a Comment