Question

The term void of x in the expression $ {{\left( x-\dfrac{3}{{{x}^{2}}} \right)}^{18}} $ is
\[\begin{align}
  & A.{}^{18}{{C}_{6}} \\
 & B.{}^{18}{{C}_{6}}{{3}^{6}} \\
 & C.{}^{18}{{C}_{5}} \\
 & D.{}^{18}{{C}_{6}}{{3}^{12}} \\
\end{align}\]

Answer 1

Hint: In this question, we need to find the term void of x in the expression of $ {{\left( x-\dfrac{3}{{{x}^{2}}} \right)}^{18}} $ which means we need to find the term which has no x. For this, we will first find the general term of the given expression and then we will use properties of exponents to simplify the general term. General term would be in terms of r. Then we will put the value of the power of x as zero which will give us the value of r. Using this value of r in general term, we will get our required term. General term r+1 of expansion of $ {{\left( a+b \right)}^{n}} $ is given as \[{{T}_{r+1}}={}^{n}{{C}_{r}}{{\left( a \right)}^{n-r}}\cdot {{\left( b \right)}^{r}}\]. Properties of exponents that we will use are:
\[\begin{align}
  & \left( i \right){{x}^{-1}}=\dfrac{1}{x} \\
 & \left( ii \right){{x}^{m}}\cdot {{x}^{n}}={{x}^{m+n}} \\
 & \left( iii \right){{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}} \\
 & \left( iv \right){{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} \\
\end{align}\]

Complete step by step answer:
Here we are given the expression as: $ {{\left( x-\dfrac{3}{{{x}^{2}}} \right)}^{18}} $ .
Let us find the general term r+1 of the expansion of this expression. We know that, in general term r+1 of the expansion $ {{\left( a+b \right)}^{n}} $ is given by \[{{T}_{r+1}}={}^{n}{{C}_{r}}{{\left( a \right)}^{n-r}}\cdot {{\left( b \right)}^{r}}\]. Comparing, we get a = x, \[b=\dfrac{-3}{{{x}^{2}}}\] and n = 18.
So any $ {{\left( r+1 \right)}^{th}} $ term of this expression will be given by \[{{T}_{r+1}}={}^{18}{{C}_{r}}{{\left( x \right)}^{18-r}}\cdot {{\left( \dfrac{-3}{{{x}^{2}}} \right)}^{r}}\].
Now let us simplify the expression of general term.
We know that, $ {{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}} $ so we get: \[{{T}_{r+1}}={}^{18}{{C}_{r}}{{\left( x \right)}^{18-r}}\cdot \dfrac{{{\left( -3 \right)}^{r}}}{{{\left( {{x}^{2}} \right)}^{r}}}\].
We know that, $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} $ so applying on \[{{\left( {{x}^{2}} \right)}^{r}}\] we get: \[{{T}_{r+1}}={}^{18}{{C}_{r}}{{x}^{18-r}}\cdot \dfrac{{{\left( -3 \right)}^{r}}}{{{x}^{2r}}}\].
Now we know that, $ {{x}^{-1}}=\dfrac{1}{x} $ so we get; \[{{T}_{r+1}}={}^{18}{{C}_{r}}{{x}^{18-r}}{{\left( -3 \right)}^{r}}{{x}^{-2r}}\].
Rearranging we get: \[{{T}_{r+1}}={}^{18}{{C}_{r}}{{\left( -3 \right)}^{r}}{{x}^{18-r}}\cdot {{x}^{-2r}}\].
We know that $ {{x}^{m}}\cdot {{x}^{n}}={{x}^{m+n}} $ so we get; \[{{T}_{r+1}}={}^{18}{{C}_{r}}{{\left( -3 \right)}^{r}}{{x}^{18-r-2r}}\Rightarrow {{T}_{r+1}}={}^{18}{{C}_{r}}{{\left( -3 \right)}^{r}}{{x}^{18-3r}}\].
Now we need the term which does not have x in it. So, we want the power of x to be zero because $ {{x}^{0}}=1 $. Hence putting the power of x as zero, we get 18-3r=0.
Simplifying this we get: $ 18=3r $ .
Dividing both sides by 3 we get: $ r=\dfrac{18}{3}=6 $ .
Hence required term will be $ {{\left( r+1 \right)}^{th}} $ term which will be $ {{7}^{th}} $ term.
So putting the value of r in the general term we get:
\[{{T}_{7}}={}^{18}{{C}_{6}}{{\left( -3 \right)}^{6}}{{x}^{0}}\Rightarrow {{T}_{7}}={}^{18}{{C}_{6}}{{\left( -3 \right)}^{6}}\].
Since power of -3 is even, so negative signs cancels out and we get: \[{{T}_{7}}={}^{18}{{C}_{6}}{{3}^{6}}\].
Hence our required term void of x is \[{}^{18}{{C}_{6}}{{3}^{6}}\].
Hence option B is the correct answer.

Note:
 Students should keep in mind the formula of a general term for an expansion of the expression $ {{\left( a+b \right)}^{n}} $ to solve this sum. They should not try to expand the expression as it will be difficult. Make sure to simplify the expression in terms of x before putting its power as zero. Take care of signs while solving.