JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \(\mathrm{m}\) and \(\mathrm{n}\) be the coefficients of seventh and thirteenth terms respectively in the expansion of \(\left(\frac{1}{3} \mathrm{x}^{\frac{1}{3}}+\frac{1}{2 \mathrm{x}^{\frac{2}{3}}}\right)^{18}\). Then \(\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}\) is :
- A \(\frac{4}{9}\)
- B \(\frac{1}{9}\)
- C \(\frac{1}{4}\)
- D \(\frac{9}{4}\)
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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