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Question: If the coefficients of \[{x^{ - 2}}\] and \({x^{ - 4}}\) in the expansion of \({\left( {{x^{\dfrac{1...

If the coefficients of x−2{x^{ - 2}} and x−4{x^{ - 4}} in the expansion of (x13+12x13)18,(x>0),{\left( {{x^{\dfrac{1}{3}}} + \dfrac{1}{{2{x^{\dfrac{1}{3}}}}}} \right)^{18}},\left( {x > 0} \right), are m and n respectively, then mn\dfrac{m}{n} is equal to :
A) 2727
B) 182182
C) 54\dfrac{5}{4}
D) 45\dfrac{4}{5}

Explanation

Solution

In this question we have to use the expansion of binomial formula , that is Tr+1=18Cr(x13)18−r(12x13)r{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 - 2 and −4 - 4 now we get the value of r put it in the equation . Now for the coefficient put x=1x = 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 - 2 and −4 - 4.
Hence we know that ,
(a+b)n{(a + b)^n} (r+1)(r + 1) term is equal to nCran−rbr{}^n{C_r}{a^{n - r}}{b^r} .
Hence for the given question (x13+12x13)18,(x>0),{\left( {{x^{\dfrac{1}{3}}} + \dfrac{1}{{2{x^{\dfrac{1}{3}}}}}} \right)^{18}},\left( {x > 0} \right),
(r+1)(r + 1) term is equal Tr+1=18Cr(x13)18−r(12x13)r{T_{r + 1}} = {}^{18}{C_r}{\left( {{x^{\dfrac{1}{3}}}} \right)^{18 - r}}{\left( {\dfrac{1}{{2{x^{\dfrac{1}{3}}}}}} \right)^r}
=18Cr(x13)18−2r(12)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{x^{ - 2}} and x−4{x^{ - 4}}
Therefore 18−2r3=−2\dfrac{{18 - 2r}}{3} = - 2 and 18−2r3=−4\dfrac{{18 - 2r}}{3} = - 4
Hence after solving we get r=12,15r = 12,15 .
Now put r=12r = 12 and x=1x = 1 for the coefficient of x−2{x^{ - 2}} that is equal to m .
m=18C12(12)12m = {}^{18}{C_{12}}{\left( {\dfrac{1}{2}} \right)^{12}}
Now put r=15r = 15 and x=1x = 1 for the coefficient of x−4{x^{ - 4}} that is equal to n .
n=18C15(12)15n = {}^{18}{C_{15}}{\left( {\dfrac{1}{2}} \right)^{15}}
Now we have to find mn\dfrac{m}{n}.
mn=18C12(12)1218C15(12)15\dfrac{m}{n} = \dfrac{{{}^{18}{C_{12}}{{\left( {\dfrac{1}{2}} \right)}^{12}}}}{{{}^{18}{C_{15}}{{\left( {\dfrac{1}{2}} \right)}^{15}}}}
Cancelling the 12\dfrac{1}{2} terms
mn=18C1218C15(12)3\dfrac{m}{n} = \dfrac{{{}^{18}{C_{12}}}}{{{}^{18}{C_{15}}{{\left( {\dfrac{1}{2}} \right)}^3}}}
mn=18C122318C15\dfrac{m}{n} = \dfrac{{{}^{18}{C_{12}}{2^3}}}{{{}^{18}{C_{15}}}}
Expansion of nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
mn=18!12!(18−12)!2318!15!(18−15)!\dfrac{m}{n} = \dfrac{{\dfrac{{18!}}{{12!\left( {18 - 12} \right)!}}{2^3}}}{{\dfrac{{18!}}{{15!\left( {18 - 15} \right)!}}}}
mn=18!12!6!2318!15!3!\dfrac{m}{n} = \dfrac{{\dfrac{{18!}}{{12!6!}}{2^3}}}{{\dfrac{{18!}}{{15!3!}}}}
Hence 18!18! present in both numerator and denominator hence it will cancel out
mn=15!3!12!6!23\dfrac{m}{n} = \dfrac{{15!3!}}{{12!6!}}{2^3}
Now expand the factorial .
mn=15.14.136.5.423\dfrac{m}{n} = \dfrac{{15.14.13}}{{6.5.4}}{2^3}
mn=273012023\dfrac{m}{n} = \dfrac{{2730}}{{120}}{2^3}
mn=22.75×8\dfrac{m}{n} = 22.75 \times 8
mn=182\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{x^{ - 2}} come as Tr+1=nα−mα+β{T_{r + 1}} = \dfrac{{n\alpha - m}}{{\alpha + \beta }} where m equals to the power of x . for expression (xα+1xβ)n{\left( {{x^\alpha } + \dfrac{1}{{{x^\beta }}}} \right)^n}. In this case m=−2,−4m = - 2, - 4 , n=18n = 18 α=β=13\alpha = \beta = \dfrac{1}{3} .
Always expand the factorial at the end of the solution otherwise the calculation becomes difficult.