QJEEM22Mabf9f: JEE Main 2022 Mathematics - Class 11 JEEM22Mabf9f Solution: Let the coefficients of x 1 and x 3 in the expansion of , be m…

Given, Binomial expansion

${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}}$

$\therefore$ General Term

${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {2{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( { - {1 \over {{x^{{1 \over 5}}}}}} \right)^r}$

$= {}^{15}{C_r}\,.\,{2^{15 - r}}\,.\,{x^{{1 \over 5}(15 - r - r)}}\,.\,{( - 1)^r}$

For ${x^{ - 1}}$ term;

${1 \over 5}(15 - 2r) = - 1$

$\Rightarrow 15 - 2r = - 5$

$\Rightarrow 2r = 20$

$\Rightarrow r = 10$

m is the coefficient of ${x^{ - 1}}$ term,

$\therefore$ $m = {}^{15}{C_{10}}\,.\,{2^{15 - 10}}\,.\,{( - 1)^{10}}$

$= {}^{15}{C_{10}}\,.\,{2^5}$

For ${x^{ - 3}}$ term;

${1 \over 5}(15 - 2r) = - 3$

$\Rightarrow 15 - 2r = - 15$

$\Rightarrow 2r = 30$

$\Rightarrow r = 15$

n is the coefficient of ${x^{ - 3}}$ term,

$\therefore$ $n = {}^{15}{C_{15}}\,.\,{2^{15 - 15}}\,.\,{( - 1)^{15}}$

$= 1\,.\,1\,.\, - 1$

$= - 1$

Given,

$m{n^2} = {}^{15}{C_r}\,.\,{2^r}$

$\Rightarrow {}^{15}{C_{10}}\,.\,{2^5}\,.\,{(1)^2} = {}^{15}{C_r}\,.\,{2^r}$ [putting value of m and n]

$\Rightarrow {}^{15}{C_{15 - 10}}\,.\,{2^5} = {}^{15}{C_r}\,.\,{2^r}$

$\Rightarrow {}^{15}{C_5}\,.\,{2^5} = {}^{15}{C_r}.\,{2^r}$

Comparing both side, we get

$r = 5$ .