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

\(\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}\) ના વિસ્તરણમાં સાતમા અને તેરમા પદ્દોના સહગુણકો અનુક્રમે \(\mathrm{m}\) અને \(\mathrm{n}\) છે. તો \(\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}} =\) ...........

  1. A  \(\frac{4}{9}\)
  2. B  \(\frac{1}{9}\)
  3. C \(\frac{1}{4}\)
  4. D  \(\frac{9}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(D)  \(\frac{9}{4}\)

Step-by-step Solution

Detailed explanation

\( \left(\frac{x^{\frac{1}{3}}}{3}+2 x^{\frac{-2}{3}}\right)^{18} \) \( t_7={ }^{18} c_6\left(\frac{x^{\frac{1}{3}}}{3}\right)^{12}\left(\frac{x^{\frac{-2}{3}}}{2}\right)^6={ }^{18} c_6 \frac{1}{(3)^{12}} \cdot \frac{1}{2^6} \)…
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