Question

If \[f(x) = {x^n}\] then the value of \[f(x)\] is \[f(1) - \left[ {\dfrac{{f'(1)}}{{1!}}} \right] + \left[ {\dfrac{{f''\left( 1 \right)}}{{2!}}} \right] - \ldots .. + \left[ {\dfrac{{{{\left( { - 1} \right)}^n}\;{f^n}\;\left( 1 \right)}}{{n!}}} \right]\] .
A. \[{{\bf{2}}^{\bf{n}}}\]
B. \[{{\bf{2}}^{{\bf{n}} - {\bf{1}}}}\]
C. 0
D. 1

Answer 1

Hint: We will find the values of given terms in the numerator section of the right side of the equal to sign and then substitute the derived values, finally at the end we will use the binomial expansion formula and reduce our derived equation to reach the final answer.

Complete step by step solution:
Firstly, here we are given
\[f(x) = {x^n}\] which signifies
\[f(1) = 1\]
Secondly, we calculate the value of f dash
\[{f^\prime }(x) = n{x^{n - 1}}\]
Next, we find the value of f double dash
\[{f^{\prime \prime }}(x) = n(n - 1){x^2}\]
Subsequently, in similar manner we can calculate the value of \[{{\rm{f}}^{\rm{n}}}({\rm{x}})\], which would be
\[{f^n}(x) = n(n - 1)(n - 2) \ldots .2.1\]
Now putting the value of X as one we get
\[{f^n}(1) = n(n - 1)(n - 2) \ldots ..2.1\]
And with further calculation
\[ = 1 - [\dfrac{n}{{1!}}] + [\dfrac{{n(n - 1)]}}{{2!}} - \dfrac{{[n(n - 1)(n - 2)]}}{{3!}} + \ldots \ldots + \dfrac{{\left[ {{{( - 1)}^n}n(n - 1)(n - 2) \ldots 2.1} \right]}}{{n!}}\]
Now by using basic binomial expansion formula \[{(1 - {\rm{x}})^{\rm{n}}}{ = ^{\rm{n}}}{{\rm{C}}_0}{1^{\rm{n}}}{( - {\rm{x}})^0}{ + ^{\rm{n}}}{{\rm{C}}_1}{1^{{\rm{n}} - 1}}{( - {\rm{x}})^1} + \ldots ..{ + ^{\rm{n}}}{C_{\rm{n}}}{( - {\rm{x}})^{\rm{n}}}\]
In the derived equation, we can reduce the equation to
\[ = {(1 - 1)^n}\]
\[ = {0^n}\]
\[ = 0\]
  Option C is the correct option.

Note: Students got confused after substitution of the differentiating values in this expression. While we use binomial expansion then we need to take of combination part.
We need to take care that combination expansion is added or not.
Combination expansion, \[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\] , where \[n\] and \[r\] both are integers.