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

Hint: 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.