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
- A \(3\)
- B \(4\)
- C \(5\)
- D \(6\)
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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