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Question: Answered & Verified by Expert
If $f(x)=x^n$, then the value of $f(1)-\frac{f^{\prime}(1)}{1}+\frac{f^{\prime \prime}(1)}{2 !}-\frac{f^{\prime \prime}(1)}{3 !}+\ldots \ldots \ldots \ldots . \frac{(-1)^n f^{\prime \prime}(1)}{n !}$ is
MathematicsSequences and SeriesJEE MainJEE Main 2003
Options:
  • A
    1
  • B
    $2^n$
  • C
    $2^{\mathrm{n}}-1$
  • D
    0
Solution:
1375 Upvotes Verified Answer
The correct answer is:
0
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}} \Rightarrow \mathrm{f}(1)=1$
$\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{nx} x^{\mathrm{n}-1} \Rightarrow \mathrm{f}^{\prime}(1)=\mathrm{n}$
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{n}(\mathrm{n}-1) \mathrm{x}^{\mathrm{n}-2} \Rightarrow \mathrm{f}^{\prime \prime}(1)=\mathrm{n}(\mathrm{n}-1)$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \mathrm{f}^{\mathrm{n}}(\mathrm{x})=\mathrm{n} ! \Rightarrow \mathrm{f}^{\mathrm{n}}(1)=\mathrm{n} !$
$=1-\frac{\mathrm{n}}{1 !}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !}+\ldots \ldots \ldots+(-1)^{\mathrm{n}} \frac{\mathrm{n} !}{\mathrm{n} !}$
$={ }^{\mathrm{n}} \mathrm{C}_0-{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2-{ }^{\mathrm{n}} \mathrm{C}_3+\ldots \ldots \ldots+(-1)^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}=0$

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