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

2ⁿ

$f(0)=1, f'(x)=n(1+x)^{n-1}, $

$f''(x)=n(n-1)(1+x)^{n-2},......,$

$f^{(n)}(x)=n(n-1) ..... 1=n!$

$\Rightarrow f'(0)=n, f''(0)=n(n-1), ....., f^n(0)=n!$

∴  Given expression = $1+\frac{n}{1}+\frac{n(n-1)}{2 !}+.....+\frac{n !}{n !}$

$={ }^n C_0+{ }^n C_1+{ }^n C_2+.....+{ }^n C_n=2^n$

Hence (2) is correct answer.