If f(x) = x^n,then the value of f(1) - frac{f | vedclass

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(C) Given $f(x) = x^n$. We know that the $k$-th derivative of $f(x)$ at $x=1$ is $f^k(1) = n(n-1)(n-2)...(n-k+1) = \frac{n!}{(n-k)!}$. Substituting th

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$0$

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(C) Given $f(x) = x^n$. We know that the $k$-th derivative of $f(x)$ at $x=1$ is $f^k(1) = n(n-1)(n-2)...(n-k+1) = \frac{n!}{(n-k)!}$. Substituting this into the expression: $f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + ...... + \frac{(-1)^n f^n(1)}{n!} = \sum_{k=0}^{n} (-1)^k \frac{f^k(1)}{k!}$. Since $\frac{f^k(1)}{k!} = \frac{n!}{k!(n-k)!} = \binom{n}{k}$,the expression becomes: $\sum_{k=0}^{n} (-1)^k \binom{n}{k} = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - ...... + (-1)^n \binom{n}{n}$. Using the binomial theorem,$(1-x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k$. Setting $x=1$,we get $(1-1)^n = 0^n = 0$ for $n \ge 1$. Thus,the value is $0$.

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

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