Let the coefficients of x^$-$1 and x^$-$3 in the expansion of ${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}},x > 0$, be m and n respectively. If r is a positive integer such that $m{n^2} = {}^{15}{C_r}\,.\,{2^r}$, then the value of r is equal to __________.

<p>Given, Binomial expansion</p> <p>${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}}$</p> <p>$\therefore$ General Term</p> <p>$${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {2{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( { - {1 \over {{x^{{1 \over 5}}}}}} \right)^r}$$</p> <p>$= {}^{15}{C_r}\,.\,{2^{15 - r}}\,.\,{x^{{1 \over 5}(15 - r - r)}}\,.\,{( - 1)^r}$</p> <p>For ${x^{ - 1}}$ term;</p> <p>${1 \over 5}(15 - 2r) = - 1$</p> <p>$\Rightarrow 15 - 2r = - 5$</p> <p>$\Rightarrow 2r = 20$</p> <p>$\Rightarrow r = 10$</p> <p>m is the coefficient of ${x^{ - 1}}$ term,</p> <p>$\therefore$ $m = {}^{15}{C_{10}}\,.\,{2^{15 - 10}}\,.\,{( - 1)^{10}}$</p> <p>$= {}^{15}{C_{10}}\,.\,{2^5}$</p> <p>For ${x^{ - 3}}$ term;</p> <p>${1 \over 5}(15 - 2r) = - 3$</p> <p>$\Rightarrow 15 - 2r = - 15$</p> <p>$\Rightarrow 2r = 30$</p> <p>$\Rightarrow r = 15$</p> <p>n is the coefficient of ${x^{ - 3}}$ term,</p> <p>$\therefore$ $n = {}^{15}{C_{15}}\,.\,{2^{15 - 15}}\,.\,{( - 1)^{15}}$</p> <p>$= 1\,.\,1\,.\, - 1$</p> <p>$= - 1$</p> <p>Given,</p> <p>$m{n^2} = {}^{15}{C_r}\,.\,{2^r}$</p> <p>$\Rightarrow {}^{15}{C_{10}}\,.\,{2^5}\,.\,{(1)^2} = {}^{15}{C_r}\,.\,{2^r}$ [putting value of m and n]</p> <p>$\Rightarrow {}^{15}{C_{15 - 10}}\,.\,{2^5} = {}^{15}{C_r}\,.\,{2^r}$</p> <p>$\Rightarrow {}^{15}{C_5}\,.\,{2^5} = {}^{15}{C_r}.\,{2^r}$</p> <p>Comparing both side, we get</p> <p>$r = 5$.</p>