Question

If \(y = [(x+1)(2x+1)(3x+1)\ldots \ldots (nx+1)]^4\), where \(n\in N\) and \(\frac{dy}{dx}\) at \(x = 0\) is \(2k\), then the value of \(k\) is

• A. \(\frac{n(n+1)}{2}\)
• B. \(n(n+1)\)
• C. \(2n(n+1)\)
• D. \(4n(n+1)\)

Hint:

Use \(y=P^4\) and evaluate \(P(0)\) and \(P'(0)\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Let \(P(x)=\prod_{k=1}^n(kx+1)\), so \(y=P^4\).

Step 2: Key Formula or Approach
\(\dfrac{dy}{dx}=4P^3P'\).

Step 3: Detailed Explanation
\(P(0)=1\). By logarithmic differentiation, \(\dfrac{P'}{P}=\sum\dfrac{k}{kx+1}\), so \(P'(0)=\sum_{k=1}^nk=\dfrac{n(n+1)}{2}\).
\[ \left.\frac{dy}{dx}\right|_{x=0}=4\cdot1\cdot\frac{n(n+1)}{2}=2n(n+1) \]
Given that this equals \(2k\), we get \(k=n(n+1)\).

Final Answer:
\(k=n(n+1)\), option (B). \[ \boxed{n(n+1)\ \text{(B)}} \]