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JEE Mains · Maths · STD 11 - 7. binomial theoram

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

  1. A \(3\)
  2. B \(4\)
  3. C \(5\)
  4. D \(6\)
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Answer & Solution

Correct Answer

(C) \(5\)

Step-by-step Solution

Detailed explanation

\(T _{ r +1}=(-1)^{ r } \cdot{ }^{15} C _{ r } \cdot 2^{15- r } X^{ \frac{15-2 r }{5}}\) \(m ={ }^{15} C _{10} 2^{5}\) \(n =-1\) \(\text { so } mn ^{2}={ }^{15} C _{5} 2^{5}\)
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