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

Correct Answer: 5

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
⇒ ¹⁵C_r 2^r = ¹⁵C₅ 2⁵
⇒ r = 5