If f(x)=(1+x)n, then the value of f(0)+f'(0)+f''(0)2!+....+fn(0)n! is
n
2n
2n -1
None of these
Explanation for correct option:
Step :Finding the value of f(0)+f'(0)+f''(0)2!+....fn(0)n!
f(x)=(1+x)nf(0)=1f'(x)=n(1+x)n-1f'(0)=nf''(x)=n(n-1)(1+x)n-2f''(0)=n(n-1)f'''(x)=n(n-1)(n-2)(1+x)n-3f'''(0)=n(n-1)(n-2)fn(x)=n(n-1)(n-2)....1fn(0)=n(n-1)(n-2)....1=f(0)+f'(0)+f''(0)2!+f'''(0)3!+.....+fn(0)n!=1+n+n(n-1)2!+....n!n!=2n
Hence, the correct option is option(B).