Final answer:
The first four non-zero terms of the Taylor series representation of 1/((x+5)(x+1)) centred at 0 are 1, -2x, 12x^2, and -48x^3.
Step-by-step explanation:
The Taylor series representation of a function centred at 0 can be found by taking the derivatives of the function. In this case, we have f(x) = 1/((x+5)(x+1)).
First, let's find the first four derivatives:
- f'(x) = -2/(x+1)^3
- f''(x) = 12/(x+1)^4
- f'''(x) = -48/(x+1)^5
- f''''(x) = 240/(x+1)^6
Next, we can use these derivatives to find the terms of the Taylor series:
- T₀(x) = 1/a₀ = 1/1 = 1
- T₁(x) = T₀(x) + f'(0)(x - 0) = 1 + (-2/1)(x - 0) = 1 - 2x
- T₂(x) = T₁(x) + f''(0)(x - 0)^2 = 1 - 2x + 12/1(x - 0)^2 = 1 - 2x + 12x^2
- T₃(x) = T₂(x) + f'''(0)(x - 0)^3 = 1 - 2x + 12x^2 - 48/1(x - 0)^3 = 1 - 2x + 12x^2 - 48x^3
Therefore, the first four non-zero terms of the Taylor series representation of 1/((x+5)(x+1)) centred at 0 are 1, -2x, 12x^2, and -48x^3.