Question

Let the coefficients of x–1 and x–3 in the expansion of
$(2x^{\frac{1}{5}} - \frac{1}{x^{\frac{1}{5}}} )^{15} , x > 0$be m and n respectively. If r is a positive integer such that
$mn² = ^{15}C_r.2^r$
then the value of r is equal to ______.

Answer 1

The correct answer is 5
Given : Expansion is $(2x^{\frac{1}{5}} - \frac{1}{x^{\frac{1}{5}}} )^{15}$
Now , the general term of given expansion is
T_{p+1} = (-1)^{p}$$^{15}C_p 2^{15-p} (x^{\frac{1}{5}})^{15-p}.(\frac{1}{x^{\frac{1}{5}}})^p
$= (-1)^p$ $^{15}C_p 2^{15-p} . x^{\frac{15-2p}{5}}$
For coefficient of $x^{-1}, \frac{15-2p}{5} = -1$
⇒ p = 10
Therefore , $m= ^{15}C_{10} 2^5$
⇒ p = 15
$∴ n = - ^{15}C_{15} 2^0 = -1$
Now , mn2 = 15C1025
⇒ mn² = 15C5 25
⇒ 15Cr 2r = 15C5 25
⇒ r = 5