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

यदि \(\left(x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{1}{3}}}\right)^{18},(x>0)\), के प्रसार में \(x^{-2}\) तथा \(x^{-4}\) के गुणांक क्रमशः \(m\) तथा \(n\) हैं, तो \(\frac{m}{n}\) बराबर है

  1. A \(27\)
  2. B \(182\)
  3. C \(\frac{5}{4}\)
  4. D \(\frac{4}{5}\)
Verified Solution

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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