If f(x)=xn, then the value of f(1)−f′(1)1!+f′′(1)2!−f′′′(1)3!+……+(−1)nfn(1)n! is

f(x)=xn⇒f(1)=1,f′(x)=nxn−1⇒f′(1)=nf′′(x)=n(n−1)xn−2⇒f′′(1)=n(n−1)….fn(x)=n!⇒fn(1)=n!∴ f(1)−f′(1)1!+f′′(1)2!……+(−1)nf′′(1)n!=1−n1!+n(n−1)2!−n(n−1)(n−2)3!+…..+(−1)nn!n!= nC0−nC1+′′C2−nC3+……+(−1)nnCn=0.

If f(x)=xn, then the value of f(1)−f′(1)1!+f′′(1)2!−f′′′(1)3!+……+(−1)nfn(1)n! is

f(x)=xn⇒f(1)=1,f′(x)=nxn−1⇒f′(1)=nf′′(x)=n(n−1)xn−2⇒f′′(1)=n(n−1)….fn(x)=n!⇒fn(1)=n!∴ f(1)−f′(1)1!+f′′(1)2!……+(−1)nf′′(1)n!=1−n1!+n(n−1)2!−n(n−1)(n−2)3!+…..+(−1)nn!n!= nC0−nC1+′′C2−nC3+……+(−1)nnCn=0.

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

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$2^{n}$

$2^{n - 1}$

answer is C.

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Detailed Solution

$f (x) = x^{n} \Rightarrow f (1) = 1, f^{'} (x) = n x^{n - 1} \Rightarrow f^{'} (1) = n$

$\begin{array}{l} f^{''} (x) = n (n - 1) x^{n - 2} \Rightarrow f^{''} (1) = n (n - 1) \dots . \\ f^{n} (x) = n! \Rightarrow f^{n} (1) = n! \end{array}$

$\begin{array}{l} ∴ f (1) - \frac{f^{'} (1)}{1!} + \frac{f^{''} (1)}{2!} \dots \dots + \frac{(- 1)^{n} f^{''} (1)}{n!} \\ = 1 - \frac{n}{1!} + \frac{n (n - 1)}{2!} - \frac{n (n - 1) (n - 2)}{3!} + \dots . . + (- 1)^{n} \frac{n!}{n!} \\ =^{n} C_{0} -^{n} C_{1} +^{''} C_{2} -^{n} C_{3} + \dots \dots + {(- 1)^{n}}^{n} C_{n} = 0 . \end{array}$