If f(x)=(1+x)n, then the value of f(0)+f'(0)+(f' '(0)/2!)+.....+(f(n)(0)/n!) is equal to

Question

If f(x)=(1+x)n, then the value of f(0)+f'(0)+(f' '(0)/2!)+.....+(f(n)(0)/n!) is equal to (A)  2n-1  (B)  2 n  (C)  n  (D)  2n

Tardigrade Admin · Accepted Answer

Given, f(x)=(1+x)n On differentiating w. r. t. x, we get f'(x)=n(1+x)n-1 Again, differentiating, we get f''(x)=n(n-1) (1+x)n-2 ∴ fn(x)=n(n-1)..... 3.2.1=n! ∴ f(0)+f'(0)+(f''(0)/2!)+....(fn(0)/n!) =1+n+(n(n-1)/2!)+....+(n!/n!) =nC0+nC1+nC2+...+nCn =2n

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

$2^{n - 1}$

$2 n$

$n$

$2^{n}$

Q. If f(x)=(1+x)n, then the value of f(0)+f′(0)+2!f′′(0)​+.....+n!f(n)(0)​ is equal to

2n−1

2n

n

2n

Given, f(x)=(1+x)n On differentiating w. r. t. x, we get f′(x)=n(1+x)n−1 Again, differentiating, we get f′′(x)=n(n−1)(1+x)n−2 ∴ fn(x)=n(n−1).....3.2.1=n! ∴ f(0)+f′(0)+2!f′′(0)​+....n!fn(0)​ =1+n+2!n(n−1)​+....+n!n!​ =nC0​+nC1​+nC2​+...+nCn​ =2n

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

$2^{n - 1}$

$2 n$

$n$

$2^{n}$