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

  • Option 1)

    2^{n}-1

  • Option 2)

    0

  • Option 3)

    1

  • Option 4)

    2^{n}

 

Answers (1)

As we learnt in 

Differentiation -

Derivative  of a function  f(x) is defined as  f'(x) means  small increment  \delta x  in x  corresponding increment in the value of y  be  \delta y

- wherein

\frac{dy}{dx}=\lim_{\delta x\rightarrow o}\:\frac{dy}{dx}

=\lim_{\delta x\rightarrow o}\:\frac{f(x+\delta x)-f(x)}{\delta x}

 

 

Since \: f(x)=x^{n}

          {f}''(x)=nx^{n-1} \: \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \: and \: so \: on

Now,

f(1)-\frac{{f}'(1)}{1!}+\frac{{f}''(1)}{2!}+\frac{{f}'''}{3!}+ \cdot \cdot \cdot \cdot \cdot \cdot

1-\frac{n}{1!}+\frac{n(n-1)}{2!}-\frac{n(n-1)(n-2)}{3!}+ \cdot \cdot \cdot \cdot \cdot \cdot \ (i)

Now,

(1+x)^{n}=1+\frac{nx}{1!}+\frac{n(n-1)x^{2}}{2!}+ \cdot \cdot \cdot \cdot \cdot \cdot

Put x=-1

0=1- \frac{n}{11}+\frac{n(n-1)}{2!} \: \cdot \cdot \cdot \cdot \cdot \cdot (ii)

From (i) and (ii) it is zero


Option 1)

2^{n}-1

This option is incorrect

Option 2)

0

This option is correct

Option 3)

1

This option is incorrect

Option 4)

2^{n}

This option is incorrect

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