<p>If \(f(x) = x^n\), then the value of \(f(1) - \dfrac{f'(1)}{1!} + \dfrac{f''(1)}{2!} - \dfrac{f'''(1)}{3!} + \cdots + \dfrac{(-1)^n f^n(1)}{n!}\) is</p>
Step-by-Step Solution
Key Concept: Recognize this as the Taylor series expansion of f(x) = x^n evaluated at x = 0, which is equivalent to the binomial expansion of (1-1)^n. The alternating sum of derivatives matches the pattern ∑(-1)^k * C(n,k) * 1^(n-k) * 1^k.
<p><strong>Step 1:</strong> Identify the general term. For f(x) = x^n, we have:</p><ul><li>f(x) = x^n</li><li>f'(x) = nx^(n-1), so f'(1) = n</li><li>f''(x) = n(n-1)x^(n-2), so f''(1) = n(n-1)</li><li>f^(k)(x) = n(n-1)...(n-k+1)x^(n-k), so f^(k)(1) = n!/(n-k)!</li></ul><p><strong>Step 2:</strong> Recognize the series. The given expression is the Taylor expansion of f(x) = x^n centered at x = 1, evaluated at x = 0:</p><p>f(1) - f'(1)/1! + f''(1)/2! - ... + (-1)^n f^(n)(1)/n! = ∑_{k=0}^{n} (-1)^k · C(n,k) · 1^(n-k) · 1^k</p><p><strong>Step 3:</strong> Apply the binomial theorem. This is the binomial expansion of (1-1)^n:</p><p>∑_{k=0}^{n} C(n,k) · 1^(n-k) · (-1)^k = (1-1)^n = 0^n</p><p><strong>Step 4:</strong> Evaluate. For n ≥ 1, we have 0^n = 0.</p><p>∴ Answer: <strong>0</strong> (Option C)</p>
Correct Answer: C