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KCET · Maths · Differentiation

If \(f(x)=1+n x+\frac{n(n-1)}{2} x^2+\frac{n(n-1)(n-2)}{6}\) \(x^3+\ldots+x^n\), then \(f^n(1)\) is equal to

  1. A \(n(n-1) 2^{n-2}\)
  2. B \(n(n-1) 2^n\)
  3. C \(2^{n-1}\)
  4. D \((n-1) 2^{n-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(n(n-1) 2^{n-2}\)

Step-by-step Solution

Detailed explanation

Given,
\(\begin{aligned} f(x)=1+n x+\frac{n(n-1)}{2} x^2+\frac{n(n-1)(n-2)}{6} x^3 & \\ & +\ldots+x^n .\end{aligned}\)
The given equation can be written as \(f(x)=(1+x)^n\)
\(\because(1+x)^n=1+n x+\frac{n(n-1)}{2 !}\)
\(x^2+\frac{n(n-1)(n-2)}{3 !} x^3+\ldots x^n\)
Now, \(f^{\prime}(x)=n(1+x)^{n-1}\)
and, \(f^{\prime \prime}(x)=n(n-1)(1+x)^{n-2}\)
put \(x=1\)
\(\therefore f^{\prime \prime}(1)=n(n-1) 2^{n-2}\)