Question

If $\mathrm{f}\left(\mathrm{x}\right)={(1+\mathrm{x})}^{\mathrm{n}}$, then the value of $\mathrm{f}\left(0\right)+\mathrm{f}\text{'}\left(0\right)+\frac{\mathrm{f}\text{'}\text{'}\left(0\right)}{2!}+....+\frac{{\mathrm{f}}^{\mathrm{n}}\left(0\right)}{\mathrm{n}!}$ is

• A. n
• B. 2ⁿ
• C. 2ⁿ -1
• D. None of these

Answer 1

The correct option is B

2ⁿ

Explanation for correct option:

Step :Finding the value of $\mathrm{f}\left(0\right)+\mathrm{f}\text{'}\left(0\right)+\frac{\mathrm{f}\text{'}\text{'}\left(0\right)}{2!}+....\frac{{\mathrm{f}}^{\mathrm{n}}\left(0\right)}{\mathrm{n}!}$

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={(1+\mathrm{x})}^{\mathrm{n}} \\ \mathrm{f}\left(0\right)=1 \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\mathrm{n}{(1+\mathrm{x})}^{\mathrm{n}-1} \\ \mathrm{f}\text{'}\left(0\right)=\mathrm{n} \\ \mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)=\mathrm{n}(\mathrm{n}-1){(1+\mathrm{x})}^{\mathrm{n}-2} \\ \mathrm{f}\text{'}\text{'}\left(0\right)=\mathrm{n}(\mathrm{n}-1) \\ \mathrm{f}\text{'}\text{'}\text{'}\left(\mathrm{x}\right)=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2){(1+\mathrm{x})}^{\mathrm{n}-3} \\ \mathrm{f}\text{'}\text{'}\text{'}\left(0\right)=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) \\ {\mathrm{f}}^{\mathrm{n}}\left(\mathrm{x}\right)=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)....1 \\ {\mathrm{f}}^{\mathrm{n}}\left(0\right)=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)....1 \\ =\mathrm{f}\left(0\right)+\mathrm{f}\text{'}\left(0\right)+\frac{\mathrm{f}\text{'}\text{'}\left(0\right)}{2!}+\frac{\mathrm{f}\text{'}\text{'}\text{'}\left(0\right)}{3!}+.....+\frac{{\mathrm{f}}^{\mathrm{n}}\left(0\right)}{\mathrm{n}!} \\ =1+\mathrm{n}+\left[\frac{\mathrm{n}(\mathrm{n}-1)}{2!}\right]+....\left(\frac{\mathrm{n}!}{\mathrm{n}!}\right) \\ =2^{\mathrm{n}} \end{gathered}$$

Hence, the correct option is option(B).