Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of $\left(\frac{1}{3}x^{\frac{1}{3}} + \frac{1}{2}x^{\frac{2}{3}}\right)^{18}$. Then $\left(\frac{n}{m}\right)^{\frac{1}{3}}$ is:

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Aicharya · Accepted Answer

Given expansion is $\left(\frac{1}{3}x^{\frac{1}{3}} + \frac{1}{2}x^{\frac{2}{3}}\right)^{18}$. $\Rightarrow T_{7} = C_{6}^{18} \left(\frac{1}{3}x^{\frac{1}{3}}\right)^{12} \left(\frac{1}{2}x^{\frac{2}{3}}\right)^{6}$ $\Rightarrow m = C_{6}^{18} \left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^{6}$ $\Rightarrow T_{13} = C_{12}^{18} \left(\frac{1}{3}x^{\frac{1}{3}}\right)^{6} \left(\frac{1}{2}x^{\frac{2}{3}}\right)^{12}$ $\Rightarrow n = C_{12}^{18} \left(\frac{1}{3}\right)^{6} \left(\frac{1}{2}\right)^{12}$ $\Rightarrow \frac{m}{n} = \frac{C_{6}^{18} \left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^{6}}{C_{12}^{18} \left(\frac{1}{3}\right)^{6} \left(\frac{1}{2}\right)^{12}}$ $\Rightarrow \frac{m}{n} = \left(\frac{1}{3}\right)^{6} \left(\frac{1}{2}\right)^{6}$ $\Rightarrow \frac{m}{n} = \left(\frac{2}{3}\right)^{6}$ $\Rightarrow \left(\frac{m}{n}\right)^{\frac{1}{3}} = \frac{4}{9}$ $\Rightarrow \left(\frac{n}{m}\right)^{\frac{1}{3}} = \frac{9}{4}$

Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of…

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