Question

If $\mathrm{f}\left(\mathrm{x}\right)=1+\mathrm{nx}+\frac{\mathrm{n}\left(\mathrm{n}-1\right)}{2!}{\mathrm{x}}^2+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3!}{\mathrm{x}}^3+....{\mathrm{x}}^{\mathrm{n}}$ ,then $f"\left(1\right)$ is equal to

• A. $n(n-1)2^{n-1}$
• B. $(n-1)2^{\mathit{n}\mathbf{-}\mathbf{1}}$
• C. $n(n-1)2^{n-2}$
• D. $n(n-1)2^n$

Answer 1

The correct option is C

$n(n-1)2^{n-2}$

Explanation for the correct options:

Binomial series:

The given equation can be written as,

$f\left(x\right)={(1+x)}^n$

$\mathbf{\because }{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}}^{\mathbf{n}}\mathbf{=}\mathbf{1}\mathbf{+}\mathit{n}\mathit{x}\mathbf{+}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{2}\mathbf{!}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{2}\mathbf{)}}{\mathbf{3}\mathbf{!}}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{x}}^{\mathbf{n}}$

Now,

$$\begin{gathered} \therefore \mathrm{f}\text{'}\left(\mathrm{x}\right)=\mathrm{n}{(1+\mathrm{x})}^{\mathrm{n}-1} \\ \mathrm{And},\mathrm{f}"\left(\mathrm{x}\right)=\mathrm{n}(\mathrm{n}-1){(1+\mathrm{x})}^{\mathrm{n}-2} \end{gathered}$$

Put $x=1,$

$\begin{array}{rcl}\therefore \mathrm{f}\text{'}\left(1\right)& =& \mathrm{n}(\mathrm{n}-1){(1+1)}^{\mathrm{n}-2}\\ & =& \mathrm{n}(\mathrm{n}-1)2^{\mathrm{n}-2}\end{array}$

Hence, the correct option is option (C)