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JEE Main 20241 FebruaryEvening Shift
Exam
JEE Main
Year
2024
Date
1 February
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2024 · 1 February · Evening Shift · Mathematics · Binomial Theorem

Let mandn be the coefficients of seventh and thirteenth terms...

Let mandnm\text{and}n be the coefficients of seventh and thirteenth terms respectively in the expansion of (13x13+12x23){\left(\dfrac{1}{3}x\dfrac{1}{3} + \dfrac{1}{2x\dfrac{2}{3}}\right)}^{18}.Then (nm)13. Then \left(\dfrac{n}{m}\right)\dfrac{1}{3} is:

Options

  1. A49\dfrac{4}{9}
  2. B19\dfrac{1}{9}
  3. C14\dfrac{1}{4}
  4. D94\dfrac{9}{4}

Correct answer

D. 94\dfrac{9}{4}

Step-by-step solution

Given expansion is (13x13+12x23)18{\left(\dfrac{1}{3}x\dfrac{1}{3} + \dfrac{1}{2{x}^{\dfrac{2}{3}}}\right)}^{18}.

⇒T7=C618(13x13)12(12x23)6⇒ {T}_{7} = C6 18 {\left(\dfrac{1}{3}x\dfrac{1}{3}\right)}^{12}{\left(\dfrac{1}{2{x}^{\dfrac{2}{3}}}\right)}^{6}

⇒m=C618(13)12(12)6⇒ m = C6 18 {\left(\dfrac{1}{3}\right)}^{12}{\left(\dfrac{1}{2}\right)}^{6}

⇒T13=C1218(13x13)6(12x23)12⇒ {T}_{13} = C12 18 {\left(\dfrac{1}{3}x\dfrac{1}{3}\right)}^{6}{\left(\dfrac{1}{2{x}^{\dfrac{2}{3}}}\right)}^{12}

⇒n=C1218(13)6(12)12⇒ n = C12 18 {\left(\dfrac{1}{3}\right)}^{6}{\left(\dfrac{1}{2}\right)}^{12}

⇒mn=C618(13)12(12)6C1218(13)6(12)12⇒ \dfrac{m}{n} = \dfrac{C6 18 {\left(\dfrac{1}{3}\right)}^{12}{\left(\dfrac{1}{2}\right)}^{6}}{C12 18 {\left(\dfrac{1}{3}\right)}^{6}{\left(\dfrac{1}{2}\right)}^{12}}

⇒mn=(13)6(12)6⇒ \dfrac{m}{n} = \dfrac{{\left(\dfrac{1}{3}\right)}^{6}}{{\left(\dfrac{1}{2}\right)}^{6}}

⇒mn=(23)6⇒ \dfrac{m}{n} = {\left(\dfrac{2}{3}\right)}^{6}

⇒(mn)13=49⇒ \left(\dfrac{m}{n}\right)\dfrac{1}{3} = \dfrac{4}{9}

⇒(nm)13=94⇒ \left(\dfrac{n}{m}\right)\dfrac{1}{3} = \dfrac{9}{4}

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