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KCET · Maths · Binomial Theorem

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

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

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Given, \(\mathrm{f}(\mathrm{x})=1+\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !} \mathrm{x}^{2}\)
\(+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !} \mathrm{x}^{3}+\ldots+\mathrm{x}^{\mathrm{n}}\)
\[
\begin{array}{ll}
\Rightarrow & \mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{\mathrm{n}} \\
\Rightarrow & \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{n}(1+\mathrm{x})^{\mathrm{n}-1} \\
\Rightarrow & \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{n}(\mathrm{n}-1)(1+\mathrm{x})^{\mathrm{n}-2} \\
\Rightarrow & \mathrm{f}^{\prime \prime}(1)=\mathrm{n}(\mathrm{n}-1) 2^{\mathrm{n}-2}
\end{array}
\]