Q101 / 180 · KCET 2023 · Mathematics
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
- Correct answerA$n(n-1) 2^{n-2}$
- B$n(n-1) 2^n$
- C$2^{n-1}$
- D$(n-1) 2^{n-1}$
Solution
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 $x=1$
$\therefore f^{\prime \prime}(1)=n(n-1) 2^{n-2}$
Hence, the correct answer is option: 'A'.