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

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:

& \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{aligned}$$ Next, we need to find the power of x in the expression $${{\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{aligned} & \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{aligned}$$ Now, the power of x is zero, when we have: $$\begin{aligned} & \Rightarrow {{x}^{\dfrac{60-5r}{6}}}={{x}^{0}} \\\ & \Rightarrow \dfrac{60-5r}{6}=0 \\\ & \Rightarrow 60-5r=0 \\\ & \Rightarrow r=12 \\\ \end{aligned}$$ So this means that $$r=12$$. Next, we know that the $$r+1$$th term of this expansion is given as: $$\begin{aligned} & \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{aligned}$$. So, we have $$r+1=13$$th term that is independent of x. Now, the value of this term when $$r=12$$ is; $$\begin{aligned} & \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{aligned}$$ Next, we know that $${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$$so we can write $$\begin{aligned} & \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{aligned}$$ So here we see that the value of the 13 th term is independent of x and is $${}^{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.