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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
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then $\mathrm{f}^{\prime \prime}(1)$ is equal to
Solution:
1631 Upvotes
Verified Answer
The correct answer is:
$n(n-1) 2^{n-2}$
Given, $\mathrm{f}(\mathrm{x})=1+\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !} \mathrm{x}^{2}$
$+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !} \mathrm{x}^{3}+\ldots+\mathrm{x}^{\mathrm{n}}$
$$
\begin{array}{ll}
\Rightarrow & \mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{\mathrm{n}} \\
\Rightarrow & \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{n}(1+\mathrm{x})^{\mathrm{n}-1} \\
\Rightarrow & \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{n}(\mathrm{n}-1)(1+\mathrm{x})^{\mathrm{n}-2} \\
\Rightarrow & \mathrm{f}^{\prime \prime}(1)=\mathrm{n}(\mathrm{n}-1) 2^{\mathrm{n}-2}
\end{array}
$$
$+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !} \mathrm{x}^{3}+\ldots+\mathrm{x}^{\mathrm{n}}$
$$
\begin{array}{ll}
\Rightarrow & \mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{\mathrm{n}} \\
\Rightarrow & \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{n}(1+\mathrm{x})^{\mathrm{n}-1} \\
\Rightarrow & \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{n}(\mathrm{n}-1)(1+\mathrm{x})^{\mathrm{n}-2} \\
\Rightarrow & \mathrm{f}^{\prime \prime}(1)=\mathrm{n}(\mathrm{n}-1) 2^{\mathrm{n}-2}
\end{array}
$$
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