Let the coefficients of x^{-1} and x^{-3} in | vedclass

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(C) The general term in the expansion of $(2x^{1/5} - x^{-1/5})^{15}$ is given by $T_{k+1} = {}^{15}C_k (2x^{1/5})^{15-k} (-x^{-1/5})^k$.$T_{k+1} = {}

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$5$

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(C) The general term in the expansion of $(2x^{1/5} - x^{-1/5})^{15}$ is given by $T_{k+1} = {}^{15}C_k (2x^{1/5})^{15-k} (-x^{-1/5})^k$.$T_{k+1} = {}^{15}C_k \cdot 2^{15-k} \cdot (-1)^k \cdot x^{(15-k)/5} \cdot x^{-k/5} = {}^{15}C_k \cdot 2^{15-k} \cdot (-1)^k \cdot x^{(15-2k)/5}$.For the coefficient of $x^{-1}$,set $(15-2k)/5 = -1 \implies 15-2k = -5 \implies 2k = 20 \implies k = 10$.So,$m = {}^{15}C_{10} \cdot 2^{15-10} \cdot (-1)^{10} = {}^{15}C_{10} \cdot 2^5 = {}^{15}C_5 \cdot 2^5$.For the coefficient of $x^{-3}$,set $(15-2k)/5 = -3 \implies 15-2k = -15 \implies 2k = 30 \implies k = 15$.So,$n = {}^{15}C_{15} \cdot 2^{15-15} \cdot (-1)^{15} = 1 \cdot 1 \cdot (-1) = -1$.Given $mn^2 = {}^{15}C_r \cdot 2^r$,we have $({}^{15}C_5 \cdot 2^5) \cdot (-1)^2 = {}^{15}C_r \cdot 2^r$.${}^{15}C_5 \cdot 2^5 = {}^{15}C_r \cdot 2^r$.Comparing both sides,we get $r = 5$.

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

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