Final answer:
To find the Taylor series for f(x) = 1/x^2 centered at a = 2, the derivatives of the function are calculated and evaluated at x = 2, showing that the series is Σ((-1)^(n+1) * (n+1) / 2^(n+2) * (x-2)^n), which is option (A).
Step-by-step explanation:
To find the Taylor series for the function f(x) = 1/x^2 centered at a = 2, we need to obtain the derivatives of the function and evaluate them at x = 2. The general formula for a Taylor series centered at a is given by:
T(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)^2/2! + ...
In this case, the function and its derivatives are:
- f(x) = 1/x^2
- f'(x) = -2/x^3
- f''(x) = 6/x^4
- f'''(x) = -24/x^5
And so on, with each derivative being evaluated at x = 2, then:
- f(2) = 1/4
- f'(2) = -1/4
- f''(2) = 3/8
- f'''(2) = -3/4
We can see a pattern in the derivatives where each term includes a factor of (-1)^{n+1}, a factorial starting from n=1, and a power of 2. Plugging these values into the Taylor series formula, we get:
Taylor series: Σ ((-1)^(n+1) * (n+1) / 2^(n+2) * (x-2)^n)
Where Σ represents the sum over n from 0 to infinity. Therefore, the correct Taylor series for f(x) = 1/x^2 centered at a = 2 is (A).