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Question: The term void of \(x\) in the expansion of \({\left( {x - \dfrac{3}{{{x^2}}}} \right)^{18}}\) is: ...

The term void of xx in the expansion of (x−3x2)18{\left( {x - \dfrac{3}{{{x^2}}}} \right)^{18}} is:
A.18C6^{18}{C_6}
B.18C636^{18}{C_6}{3^6}
C.18C5^{18}{C_5}
D.18C6312^{18}{C_6}{3^{12}}

Explanation

Solution

Hint: We want to find a term void of xx which also means a term independent of xx. Hence, find the term in the expansion where power of xx is zero. Use the formula of the general term of binomial expansion and find the power of xx. Equate it 0 to find the value of rr, substitute the value of rr in the formula of the general term to find the term void of xx.

Complete step-by-step answer:
We have to find the term in the binomial expansion of (x−3x2)18{\left( {x - \dfrac{3}{{{x^2}}}} \right)^{18}} that is void of xx or we can say, independent of xx.
Thus, we need to find the term in the expansion such that the power of xx is 0.
First of all, let us find out the general term in the binomial expansion of (x−3x2)18{\left( {x - \dfrac{3}{{{x^2}}}} \right)^{18}}.
The formula of general term is, Tr+1=nCran−rbr{T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}, where 0⩽r<n0 \leqslant r < n in the expansion of (a+b)n{\left( {a + b} \right)^n}.
On substituting a=x,b=−3x2,n=18a = x,b = - \dfrac{3}{{{x^2}}},n = 18 in the general term formula we get,
Tr+1=18Cr(x)18−r(−3x2)r{T_{r + 1}}{ = ^{18}}{C_r}{\left( x \right)^{18 - r}}{\left( { - \dfrac{3}{{{x^2}}}} \right)^r}
Combine the terms of xx.
Tr+1=18Cr(x)18−r(−3x2)r Tr+1=18Cr(x)18−r(−3)r(x)−2r Tr+1=18Cr(−3)r(x)18−3r  {T_{r + 1}}{ = ^{18}}{C_r}{\left( x \right)^{18 - r}}{\left( { - \dfrac{3}{{{x^2}}}} \right)^r} \\\ {T_{r + 1}}{ = ^{18}}{C_r}{\left( x \right)^{18 - r}}{\left( { - 3} \right)^r}{\left( x \right)^{ - 2r}} \\\ {T_{r + 1}}{ = ^{18}}{C_r}{\left( { - 3} \right)^r}{\left( x \right)^{18 - 3r}} \\\
Now, to find the term independent of xx, so put 18−3r=018 - 3r = 0 to find the value of rr.
18−3r=0 18=3r r=6  18 - 3r = 0 \\\ 18 = 3r \\\ r = 6 \\\
On substituting the value of rrin general term formula Tr+1=18Cr(−3)r(x)18−3r{T_{r + 1}}{ = ^{18}}{C_r}{\left( { - 3} \right)^r}{\left( x \right)^{18 - 3r}}, we get
T6+1=18C6(−3)6(x)18−3(6) T7=18C6(−3)6 T7=18C6(3)6  {T_{6 + 1}}{ = ^{18}}{C_6}{\left( { - 3} \right)^6}{\left( x \right)^{18 - 3\left( 6 \right)}} \\\ {T_7}{ = ^{18}}{C_6}{\left( { - 3} \right)^6} \\\ {T_7}{ = ^{18}}{C_6}{\left( 3 \right)^6} \\\
Thus, the term void of xx is 18C636^{18}{C_6}{3^6}.
Hence, B is the correct option.

Note: Any number raised to the power 0 is 1. Hence, to get a term independent of xx find the term in which the power of xx is zero. Many students do this question by expanding the expression term by term, which makes the solution unnecessarily long.