Let coefficients of x ^{-1} and x ^{-3} in expansion of left(2 x^{frac{1}{5}}-frac{1}{x^{frac{1}{5}}

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7.Binomial Theorem

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

A

$3$

B

$4$

C

$5$

D

$6$

(JEE MAIN-2022)

Solution

$T _{ r +1}=(-1)^{ r } \cdot{ }^{15} C _{ r } \cdot 2^{15- r } X^{ \frac{15-2 r }{5}}$

$m ={ }^{15} C _{10} 2^{5}$

$n =-1$

$\text { so } mn ^{2}={ }^{15} C _{5} 2^{5}$

Standard 11

Mathematics

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