JEE Mains · Maths · STD 11 - 7. binomial theoram
If the coefficients of \(x^{-2}\) and \(x^{-4}\) in the expansion of \({\left( {{x^{\frac{1}{3}}} + \frac{1}{{2{x^{\frac{1}{3}}}}}} \right)^{18}}\,,\,\left( {x > 0} \right),\) are \(m\) and \(n\) respectively, then \(\frac{m}{n}\) is equal to
- A \(27\)
- B \(182\)
- C \(\frac{5}{4}\)
- D \(\frac{4}{5}\)
Answer & Solution
Correct Answer
(B) \(182\)
Step-by-step Solution
Detailed explanation
\(T_{r+1}=18 C_{r}\left(x^{\frac{1}{3}}\right)^{18-r}\left(\frac{1}{2 x^{\frac{1}{3}}}\right)^{r}\) \(=^{18} C_{r} x^{6-\frac{2 r}{3}} \frac{1}{2^{r}}\)…
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