Question

The binomial expansion of ${\left( {{x^k} + \dfrac{1}{{{x^{2k}}}}} \right)^{3n}}$,$n \in N$, contains a term independent of $x$.
A. Only if $k$ is an integer
B. Only if $k$ is a natural number
C. Only if $k$ is rational
D. For a real $k$

Answer 1

First we will compare the general term or ${\left( {r + 1} \right)^{th}}$ term in the expansion ${\left( {a + b} \right)^n}$ is given by ${}^n{C_r}{a^{n - 1}}{b^r}$ using the binomial theorem with the given expansion to find the value of $a$ and $b$. Since we are given that the fifth term is independent of $x$, then we will find the value of $n$ such that the exponent powers of $x$ will be equal in the obtained equation to find the value of $k$.

Complete step by step answer:

We are given that the expansion is ${\left( {{x^k} + \dfrac{1}{{{x^{2k}}}}} \right)^{3n}}$.

We know that the general term or ${\left( {r + 1} \right)^{th}}$ term in the expansion ${\left( {a + b} \right)^N}$ is given by ${}^N{C_r}{a^{N - r}}{b^r}$ using the binomial theorem.
Comparing the given expansion with above expansion to find the value of $a$, $b$ and $N$.
$N = 3n$
$a = {x^k}$
$b = \dfrac{1}{{{x^{2k}}}}$
Substituting these values in the above general term, we get
$\Rightarrow {}^{3n}{C_r}{x^{\left( {3n - r} \right)k}}\dfrac{1}{{{x^{2kr}}}}$
Since we are given that the term is independent of $x$, then the exponents powers of $x$ will be equal in the above equation, we get
$\Rightarrow 2kr = \left( {3n - r} \right)k$
Dividing the above equation by $k$ on both sides, we get

$$\Rightarrow \dfrac{{2kr}}{k} = \dfrac{{\left( {3n - r} \right)k}}{k} \\\ \Rightarrow 2r = 3n - r \\\$$

Adding the above equation by $r$ on both sides, we get

$$\Rightarrow 2r + r = 3n - r + r \\\ \Rightarrow 3r = 3n \\\$$

Dividing the above equation by 3 on both sides, we get

$$\Rightarrow \dfrac{{3r}}{3} = \dfrac{{3n}}{3} \\\ \Rightarrow r = n \\\$$

Since it does not depend on $k$, so the given expansion contains a term independent of $x$ for any real $k$.
Hence, option D is correct.

Note: In solving these types of questions, some student try to open the combinations using the formula, ${}^n{C_r} = \dfrac{{\left. {\underline {\, n \,}}\\! \right| }}{{\left. {\underline {\, r \,}}\\! \right| \cdot \left. {\underline {\, {n - r} \,}}\\! \right| }}$, where $n$ is the number of items, and $r$ represents the number of items being chosen, which is wrong. So, we have to just take the exponential powers of $x$ to find the required value, instead of opening and solving the combinations, or else it will confuse the student. Do not write $n$ instead of $3n$ will lead to the wrong answer.