Question

If f(x) = 1 + nx + \(\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}x^3+...+x^n\) then f"(1) =

• A. n(n - 1) 2ⁿ
• B. (n - 1) 2^n-1
• C. 2^n-1
• D. n(n - 1)2^n-2

Hint:

When differentiating polynomials, take the derivative term by term. In this case, the second derivative of the polynomial involves applying the power rule multiple times. For polynomials of the form \( f(x) = 1 + nx + \frac{n(n-1)}{2}x^2 + ... \), the general form of the second derivative at \( x = 1 \) simplifies to the leading term \( n(n - 1) 2^{n-2} \). This is useful for handling higher-order derivatives of polynomial functions.

Answer 1

The Correct Option is D

The correct answer is (D) : n(n - 1)2^n-2.

Answer 2

The correct answer is: (D) \( n(n - 1) 2^{n-2} \).

We are given the function \( f(x) \):

\(f(x) = 1 + nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{6}x^3 + ... + x^n\)

We are tasked with finding \( f''(1) \), the second derivative of \( f(x) \) evaluated at \( x = 1 \). 
Step 1: Differentiate \( f(x) \)

The given expression for \( f(x) \) is a polynomial, and its general form is a sum of terms involving powers of \( x \). The second derivative of a polynomial is simply the term-by-term derivative taken twice. The first derivative \( f'(x) \) is: \[ f'(x) = n + \frac{n(n-1)}{2} \cdot 2x + \frac{n(n-1)(n-2)}{6} \cdot 3x^2 + ... + n \cdot x^{n-1} \] Simplifying the terms: \[ f'(x) = n + n(n-1) x + n(n-1)(n-2) x^2 + ... + n x^{n-1} \] Now, differentiate \( f'(x) \) to get \( f''(x) \): \[ f''(x) = n(n-1) + 2n(n-1)(n-2) x + 3n(n-1)(n-2)(n-3) x^2 + ... + n(n-1)(n-2) x^{n-3} \] Step 2: Evaluate \( f''(x) \) at \( x = 1 \)

Substituting \( x = 1 \) into the expression for \( f''(x) \): \[ f''(1) = n(n-1) + 2n(n-1)(n-2) + 3n(n-1)(n-2)(n-3) + ... + n(n-1)(n-2) \] The highest degree term is \( n(n-1) 2^{n-2} \), and this is the value of \( f''(1) \).
Therefore, the correct answer is (D) \( n(n - 1) 2^{n-2} \).