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माना $\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}$ के प्रसार में सातवें तथा तेरहवें पदों के गुणांक क्रमशः $m$ तथा $n$ है। तो $\left(\frac{n}{m}\right)^{\frac{1}{3}}$ बराबर है :
$\frac{4}{9}$
$\frac{1}{9}$
$\frac{1}{4}$
$\frac{9}{4}$
Solution
$ \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} $
$ \mathrm{t}_{13}={ }^{18} c_{12}\left(\frac{x^{\frac{1}{3}}}{3}\right)^6\left(\frac{x^{\frac{-2}{3}}}{2}\right)^{12}={ }^{18} c_{12} \frac{1}{(3)^6} \cdot \frac{1}{2^{12}} \cdot x^{-6} $
$ \mathrm{~m}={ }^{18} \mathrm{c}_6 \cdot 3^{-12} \cdot 2^{-6}: \mathrm{n}={ }^{18} \mathrm{c}_{12} \cdot 2^{-12} \cdot 3^{-6} $
$ \left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}=\left(\frac{2^{-12} \cdot 3^{-6}}{3^{-12} \cdot 2^{-6}}\right)^{\frac{1}{3}}=\left(\frac{3}{2}\right)^2=\frac{9}{4}$
