If f(x)=xn+4, then the value of f(1)+(f prime(1)/1 !)+(f prime prime(1)/2 !)+(f prime prime prime(1)/3 !)+ ldots+(fn(1)/n !) is

Question

If f(x)=xn+4, then the value of f(1)+(f prime(1)/1 !)+(f prime prime(1)/2 !)+(f prime prime prime(1)/3 !)+ ldots+(fn(1)/n !) is (A) 2n-1 (B) 2n+4 (C) 1+(1/1 !)+(1/2 !)+ ldots+(1/n !) (D) None of these

Tardigrade Admin · Accepted Answer

f(1)=5, f prime(x)=n xn-1, f prime prime(x)=n(n-1) xn-2 So, f prime(1)=n f prime prime(1)=n(n-1) ldots fn(1)=12 ldots n Thus, f(1)+(f prime(1)/1 !)+ ldots+(fn(1)/n !) =5+(n/1)+(n(n-1)/2 !)+ ldots+(n !/n !) =(1+1)n+4=2n+4

Q. If $f (x) = x^{n} + 4$ , then the value of $f (1) + \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} + \frac{f ^{'''} ( 1 )}{3 !} + \dots + \frac{f ^{n} ( 1 )}{n !}$ is

$2^{n - 1}$

$2^{n} + 4$

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

None of these

Q. If f(x)=xn+4, then the value of f(1)+1!f′(1)​+2!f′′(1)​+3!f′′′(1)​+…+n!fn(1)​ is

2n−1

2n+4

1+1!1​+2!1​+…+n!1​

None of these

f(1)=5,f′(x)=nxn−1,f′′(x)=n(n−1)xn−2 So, f′(1)=n f′′(1)=n(n−1)…fn(1)=12…n Thus, 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 $f (1) + \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} + \frac{f ^{'''} ( 1 )}{3 !} + \dots + \frac{f ^{n} ( 1 )}{n !}$ is

$2^{n - 1}$

$2^{n} + 4$

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