Question

Find the term independent of x in\[{{\left( 2{{x}^{\dfrac{1}{2}}}-3{{x}^{-\dfrac{1}{3}}} \right)}^{20}}\]. Choose the correct option,
A. \[{}^{20}{{C}_{8}}\bullet {{6}^{8}}\bullet {{2}^{4}}\]
B. \[{}^{20}{{C}_{8}}\bullet {{2}^{8}}\bullet {{3}^{8}}\]
C. \[{}^{20}{{C}_{8}}\bullet {{6}^{8}}\bullet {{3}^{4}}\]
D. \[{}^{20}{{C}_{12}}\bullet {{6}^{12}}\]

Answer 1

Hint: Expand the given expression using the expansion formula\[{{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{a}^{\left( n-r \right)}}{{b}^{r}}\]. Here \[a=2{{x}^{\dfrac{1}{2}}},\,\,b=3{{x}^{-\dfrac{1}{3}}}\]and \[n=20\]. Next, we know that the \[r+1\]th term of this expansion is given as: \[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{\left( n-r \right)}}{{b}^{r}}\]. So here we have to find the total power of x in terms of r, and then make this total power zero (i.e, \[{{x}^{0}}\] ) in order to find the value of r. Then we can find the term and the value of that term which is independent of x.

Complete step-by-step answer:
In the question, we have to find the term independent of x in the expansion of \[{{\left( 2{{x}^{\dfrac{1}{2}}}-3{{x}^{-\dfrac{1}{3}}} \right)}^{20}}\].
So here we can use the binomial expansion. The binomial expansion of expression of the form \[{{\left( a+b \right)}^{n}}\]is given as\[{{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{a}^{\left( n-r \right)}}{{b}^{r}}\]. Here a and b can be a variable or a constant. And the power n can be an integer or a fraction. Now, the expression that we have is \[{{\left( 2{{x}^{\dfrac{1}{2}}}-3{{x}^{-\dfrac{1}{3}}} \right)}^{20}}\]. So on comparing with the form \[{{\left( a+b \right)}^{n}}\] we get \[a=2{{x}^{\dfrac{1}{2}}},\,\,b=3{{x}^{-\dfrac{1}{3}}}\]and \[n=20\]. So now when we expand the expression, we have:
\[\begin{align}
  & \Rightarrow {{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{a}^{\left( n-r \right)}}{{b}^{r}} \\
 & \Rightarrow {{\left( 2{{x}^{\dfrac{1}{2}}}-3{{x}^{-\dfrac{1}{3}}} \right)}^{20}}=\sum\limits_{r=0}^{20}{{}^{20}{{C}_{r}}}{{\left( 2{{x}^{\dfrac{1}{2}}} \right)}^{\left( 20-r \right)}}{{\left( 3{{x}^{-\dfrac{1}{3}}} \right)}^{r}} \\
\end{align}\]
Next, we need to find the power of x in the expression \[{{\left( 2{{x}^{\dfrac{1}{2}}} \right)}^{\left( 20-r \right)}}{{\left( 3{{x}^{-\dfrac{1}{3}}} \right)}^{r}}\]. So that is calculated as follows:
\[\begin{align}
  & \Rightarrow \left( {{x}^{\dfrac{20-r}{2}}} \right)\left( {{x}^{-\dfrac{r}{3}}} \right) \\
 & \Rightarrow {{x}^{\dfrac{20-r}{2}-\dfrac{r}{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because {{x}^{a}}\times {{x}^{b}}={{x}^{a+b}} \\
 & \Rightarrow {{x}^{\dfrac{60-3r-2r}{6}}} \\
 & \Rightarrow {{x}^{\dfrac{60-5r}{6}}} \\
\end{align}\]
Now, the power of x is zero, when we have:
\[\begin{align}
  & \Rightarrow {{x}^{\dfrac{60-5r}{6}}}={{x}^{0}} \\
 & \Rightarrow \dfrac{60-5r}{6}=0 \\
 & \Rightarrow 60-5r=0 \\
 & \Rightarrow r=12 \\
\end{align}\]
So this means that \[r=12\]. Next, we know that the \[r+1\]th term of this expansion is given as: \[\begin{align}
  & \Rightarrow {{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{\left( n-r \right)}}{{b}^{r}} \\
 & \Rightarrow {{T}_{r+1}}={}^{20}{{C}_{r}}{{\left( 2{{x}^{\dfrac{1}{2}}} \right)}^{\left( 20-r \right)}}{{\left( 3{{x}^{-\dfrac{1}{3}}} \right)}^{r}} \\
\end{align}\].
So, we have \[r+1=13\]th term that is independent of x. Now, the value of this term when \[r=12\] is;
\[\begin{align}
  & \Rightarrow {{T}_{12+1}}={}^{20}{{C}_{12}}{{\left( 2{{x}^{\dfrac{1}{2}}} \right)}^{\left( 20-12 \right)}}{{\left( 3{{x}^{-\dfrac{1}{3}}} \right)}^{12}} \\
 & \Rightarrow {{T}_{13}}={}^{20}{{C}_{12}}{{\left( 2 \right)}^{\left( 20-12 \right)}}{{\left( 3 \right)}^{12}}{{x}^{0}} \\
 & \Rightarrow {{T}_{13}}={}^{20}{{C}_{12}}{{\left( 2 \right)}^{\left( 8 \right)}}{{\left( 3 \right)}^{12}} \\
\end{align}\]
Next, we know that \[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\]so we can write
\[\begin{align}
  & \Rightarrow {{T}_{13}}={}^{20}{{C}_{12}}{{\left( 2 \right)}^{\left( 8 \right)}}{{\left( 3 \right)}^{12}} \\
 & \Rightarrow {{T}_{13}}={}^{20}{{C}_{20-12}}{{\left( 2 \right)}^{\left( 8 \right)}}{{\left( 3 \right)}^{12}} \\
 & \Rightarrow {{T}_{13}}={}^{20}{{C}_{8}}{{\left( 6 \right)}^{\left( 8 \right)}}{{\left( 3 \right)}^{4}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because {{\left( 2 \right)}^{\left( 8 \right)}}{{\left( 3 \right)}^{\left( 8 \right)}}={{\left( 6 \right)}^{\left( 8 \right)}} \\
\end{align}\]
So here we see that the value of the 13 th term is independent of x and is \[{}^{20}{{C}_{8}}{{\left( 6 \right)}^{\left( 8 \right)}}{{\left( 3 \right)}^{4}}\]. Hence the correct answer is option C.

Note: The general term of the expansion of the form \[{{\left( a+b \right)}^{n}}\] does not represent the r th term but r+1 th term. Also, we have to be careful to find the total power and then make it zero to find the term independent of x. Also, we are applying the exponent rule \[{{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}\], then we have to be careful with the negative powers.