Question

If the coefficients of \[{x^{ - 2}}\] and ${x^{ - 4}}$ in the expansion of ${\left( {{x^{\dfrac{1}{3}}} + \dfrac{1}{{2{x^{\dfrac{1}{3}}}}}} \right)^{18}},\left( {x > 0} \right),$ are m and n respectively, then $\dfrac{m}{n}$ is equal to :
A) $27$
B) $182$
C) $\dfrac{5}{4}$
D) $\dfrac{4}{5}$

Answer 1

Hint: In this question we have to use the expansion of binomial formula , that is ${T_{r + 1}} = {}^{18}{C_r}{\left( {{x^{\dfrac{1}{3}}}} \right)^{18 - r}}{\left( {\dfrac{1}{{2{x^{\dfrac{1}{3}}}}}} \right)^r}$ after this solve it and compare the power of x equal to $ - 2$ and $ - 4$ now we get the value of r put it in the equation . Now for the coefficient put $x = 1$ get the value.

Complete step-by-step answer:
In this first we have to find the term in which the power of x is equal to $ - 2$ and $ - 4$.
Hence we know that ,
${(a + b)^n}$ $(r + 1)$ term is equal to ${}^n{C_r}{a^{n - r}}{b^r}$ .
Hence for the given question ${\left( {{x^{\dfrac{1}{3}}} + \dfrac{1}{{2{x^{\dfrac{1}{3}}}}}} \right)^{18}},\left( {x > 0} \right),$
$(r + 1)$ term is equal ${T_{r + 1}} = {}^{18}{C_r}{\left( {{x^{\dfrac{1}{3}}}} \right)^{18 - r}}{\left( {\dfrac{1}{{2{x^{\dfrac{1}{3}}}}}} \right)^r}$
$ = {}^{18}{C_r}{\left( {{x^{\dfrac{1}{3}}}} \right)^{18 - 2r}}{\left( {\dfrac{1}{2}} \right)^r}$
Now in this question we have to find the coefficient of term \[{x^{ - 2}}\] and ${x^{ - 4}}$
Therefore $\dfrac{{18 - 2r}}{3} = - 2$ and $\dfrac{{18 - 2r}}{3} = - 4$
Hence after solving we get $r = 12,15$ .
Now put $r = 12$ and $x = 1$ for the coefficient of ${x^{ - 2}}$ that is equal to m .
$m = {}^{18}{C_{12}}{\left( {\dfrac{1}{2}} \right)^{12}}$
Now put $r = 15$ and $x = 1$ for the coefficient of ${x^{ - 4}}$ that is equal to n .
$n = {}^{18}{C_{15}}{\left( {\dfrac{1}{2}} \right)^{15}}$
Now we have to find $\dfrac{m}{n}$.
$\dfrac{m}{n} = \dfrac{{{}^{18}{C_{12}}{{\left( {\dfrac{1}{2}} \right)}^{12}}}}{{{}^{18}{C_{15}}{{\left( {\dfrac{1}{2}} \right)}^{15}}}}$
Cancelling the $\dfrac{1}{2}$ terms
$\dfrac{m}{n} = \dfrac{{{}^{18}{C_{12}}}}{{{}^{18}{C_{15}}{{\left( {\dfrac{1}{2}} \right)}^3}}}$
$\dfrac{m}{n} = \dfrac{{{}^{18}{C_{12}}{2^3}}}{{{}^{18}{C_{15}}}}$
Expansion of ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$\dfrac{m}{n} = \dfrac{{\dfrac{{18!}}{{12!\left( {18 - 12} \right)!}}{2^3}}}{{\dfrac{{18!}}{{15!\left( {18 - 15} \right)!}}}}$
\[\dfrac{m}{n} = \dfrac{{\dfrac{{18!}}{{12!6!}}{2^3}}}{{\dfrac{{18!}}{{15!3!}}}}\]
Hence $18!$ present in both numerator and denominator hence it will cancel out
\[\dfrac{m}{n} = \dfrac{{15!3!}}{{12!6!}}{2^3}\]
Now expand the factorial .
\[\dfrac{m}{n} = \dfrac{{15.14.13}}{{6.5.4}}{2^3}\]
\[\dfrac{m}{n} = \dfrac{{2730}}{{120}}{2^3}\]
\[\dfrac{m}{n} = 22.75 \times 8\]
\[\dfrac{m}{n} = 182\]

Hence option B will be the correct answer.

Note: We can also use the formula for finding the term in which ${x^{ - 2}}$ come as ${T_{r + 1}} = \dfrac{{n\alpha - m}}{{\alpha + \beta }}$ where m equals to the power of x . for expression ${\left( {{x^\alpha } + \dfrac{1}{{{x^\beta }}}} \right)^n}$. In this case $m = - 2, - 4$ , $n = 18$ $\alpha = \beta = \dfrac{1}{3}$ .
 Always expand the factorial at the end of the solution otherwise the calculation becomes difficult.