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: Take logs
$\log y=4\sum\log(kx+1)$, so $\dfrac{y'}{y}=4\sum\dfrac{k}{kx+1}$.

Step 2: Evaluate at 0
$y(0)=1$ and the sum is $\dfrac{n(n+1)}{2}$, so $y'(0)=2n(n+1)=2k$ and $k=n(n+1)$, option (B).

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