Question

If $f\left( x \right) = {\left( {2011 + x} \right)^n}$, where $x$ is a real variable and $n$ is a positive integer, then value of $f\left( 0 \right) + {f'}\left( 0 \right) + \dfrac{{{f{''}}}}{{2!}} + \cdot \cdot \cdot \cdot \cdot \dfrac{{{f^{\left( {n - 1} \right)}}\left( 0 \right)}}{{\left( {n - 1} \right)}}$ is
A. ${\left( {2011} \right)^n}$
B. ${\left( {2012} \right)^n}$
C. ${\left( {2012} \right)^n} - 1$
D. $n{\left( {2011} \right)^n}$

Answer 1

Hint: Find the value of the given condition $f\left( 0 \right) + {f'}\left( 0 \right) + \dfrac{{{f{''}}}}{{2!}} + \cdot \cdot \cdot \cdot \cdot \dfrac{{{f^{\left( {n - 1} \right)}}\left( 0 \right)}}{{\left( {n - 1} \right)}}$ from binomial expansion by substituting the values in the function $f\left( x \right) = {\left( {2011 + x} \right)^n}$ and simplify the expression to get the answer.

Complete step-by-step answer:
According to the question it is given that the function is,
$f\left( x \right) = {\left( {2011 + x} \right)^n}$
The value of $f\left( 0 \right)$ can be defined by putting $0$ in place of $x$.
$f\left( 0 \right) = {\left( {2011 + 0} \right)^n}$
$f\left( 0 \right) = {\left( {2011} \right)^n}$……..(1)
Similarly, the value of ${f'}\left( 0 \right)$ can be calculated by substituting $0$ in place of $x$.
${f'}\left( 0 \right) = {}^n{C_1}{2011^{n - 1}}$……….(2)
Proceed in the same way and find the value of all expressions.
$\dfrac{{{f{''}}\left( 0 \right)}}{{2!}} = {}^n{C_2}{2011^{n - 2}}$ …………(3)
And,
$\dfrac{{{f^{n - 1}}\left( 0 \right)}}{{\left( {n - 1} \right)!}} = {}^n{C_{n - 1}}2011$
$\dfrac{{{f^{n - 1}}\left( 0 \right)}}{{\left( {n - 1} \right)!}} = 2011$……….(4)
Add the above all equations.
${\left( {2011} \right)^n} + {}^n{C_1}{\left( {2011} \right)^{n - 1}} + {}^n{C_2}{\left( {2011} \right)^{n - 2}} + \cdot \cdot \cdot \cdot \cdot {}^n{C_{n - 1}}2011$ …….(5)
Add and subtract ${}^n{C_n}$ in equation (5).
The above term ${\left( {2011} \right)^n} + {}^n{C_1}{\left( {2011} \right)^{n - 1}} + {}^n{C_2}{\left( {2011} \right)^{n - 2}} + \cdot \cdot \cdot \cdot \cdot {}^n{C_{n - 1}}2011$ becomes,
${\left( {2011 + 1} \right)^n} - 1 = {\left( {2012} \right)^n} - 1\;\;\;\;\;\left( {{}^n{C_n} = 1} \right)$

Hence, the option C is correct.

Note: Binomial expansion describes the algebraic expansion of the powers of a binomial. The binomial theorem specifies the expansion of any power. The larger the power is the harder it is to expand the expressions but binomial theorem makes it easier. Binomial means consisting of two terms
Expansion is defined as the act of getting bigger added onto something. Binomial expansion is very useful in finding the answers of any power. Binomial theorem is used in many fields.