If f(x)=xn+4 then the value of f (1)+( f prime(1)/1 !)+( f prime prime(1)/2 !)+ ldots+( f n (1)/ n !) is

Question

If f(x)=xn+4 then the value of f (1)+( f prime(1)/1 !)+( f prime prime(1)/2 !)+ ldots+( f n (1)/ n !) is (A) 2 n -1 (B) 2 n +4 (C) 1+(1/1 !)+(1/2 !)+ ldots+(1/n !) (D) 0

Tardigrade Admin · Accepted Answer

f(1)=5, f prime(x)=n xn-1, so f prime(1)=n f prime prime(1)=n(n-1) ldots then fn(1)=1.2 ldots . n f (1) +( f prime(1)/1 !)+ ldots+( f n (1)/ n !) =5+( n /1)+( n ( n -1)/2 !)+ ldots+( n !/ n !) =(1+1) n +4=2 n +4

Q. If $f (x) = x^{n} + 4$ then the value of $f (1) + \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} + \dots + \frac{f ^{n} ( 1 )}{n !}$ is

$2^{n - 1}$

$2^{n} + 4$

$1 + \frac{1}{1 !} + \frac{1}{2 !} + \dots + \frac{1}{n !}$

0

$f (1) = 5, f^{'} (x) = n x^{n - 1}$ , so $f^{'} (1) = n$
$f^{''} (1) = n (n - 1) \dots$ then $f^{n} (1) = 1.2 \dots . n$
$f (1) + \frac{f ^{'} ( 1 )}{1 !} + \dots + \frac{f ^{n} ( 1 )}{n !}$
$= 5 + \frac{n}{1} + \frac{n ( n - 1 )}{2 !} + \dots + \frac{n !}{n !}$
$= (1 + 1)^{n} + 4 = 2^{n} + 4$

Q. If f(x)=xn+4 then the value of f(1)+1!f′(1)​+2!f′′(1)​+…+n!fn(1)​ is

2n−1

2n+4

1+1!1​+2!1​+…+n!1​

0

f(1)=5,f′(x)=nxn−1, so f′(1)=n f′′(1)=n(n−1)… then fn(1)=1.2….n f(1)+1!f′(1)​+…+n!fn(1)​ =5+1n​+2!n(n−1)​+…+n!n!​ =(1+1)n+4=2n+4

Q. If $f (x) = x^{n} + 4$ then the value of $f (1) + \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} + \dots + \frac{f ^{n} ( 1 )}{n !}$ is

$2^{n - 1}$

$2^{n} + 4$

$1 + \frac{1}{1 !} + \frac{1}{2 !} + \dots + \frac{1}{n !}$

$f (1) = 5, f^{'} (x) = n x^{n - 1}$ , so $f^{'} (1) = n$
$f^{''} (1) = n (n - 1) \dots$ then $f^{n} (1) = 1.2 \dots . n$
$f (1) + \frac{f ^{'} ( 1 )}{1 !} + \dots + \frac{f ^{n} ( 1 )}{n !}$
$= 5 + \frac{n}{1} + \frac{n ( n - 1 )}{2 !} + \dots + \frac{n !}{n !}$
$= (1 + 1)^{n} + 4 = 2^{n} + 4$