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

If f(x)=xn,f(x) = x^{n}, then the value of

f(1)−f′(1)1!+f′′(1)2!−f′′′(1)3!+......+(−1)nfn(1)n!f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + ...... + \frac{( - 1)^{n}f^{n}(1)}{n!} is

A

2n2^{n}

B

2n−12^{n - 1}

C

0

D

1

Answer

0

Explanation

Solution

f(x)=xn⇒f(1)=1f(x) = x^{n} \Rightarrow f(1) = 1, f′(x)=nxn−1⇒f′(1)=nf^{'}(x) = nx^{n - 1} \Rightarrow f^{'}(1) = n

f′′(x)=n(n−1)xn−2⇒f′′(1)=n(n−1)f^{''}(x) = n(n - 1)x^{n - 2} \Rightarrow f^{''}(1) = n(n - 1) …..

fn(x)=n!⇒fn(1)=n!f^{n}(x) = n! \Rightarrow f^{n}(1) = n!, ∴f(1)−f′(1)1!+f′′(1)2!......+(−1)nfn(1)n!\therefore f(1) - \frac{f^{'}(1)}{1!} + \frac{f^{''}(1)}{2!}...... + \frac{( - 1)^{n}f^{n}(1)}{n!}

=n⥂C0−n⥂C1+n⥂C2−n⥂C3+......+(−1)nn⥂Cn=0=^{n} ⥂ C_{0} -^{n} ⥂ C_{1} +^{n} ⥂ C_{2} -^{n} ⥂ C_{3} + ...... + ( - 1{)^{n}}^{n} ⥂ C_{n} = 0.