Let m andn be the coefficients of seventh and thirteenth terms respectively in the expansion of ( 1 3 x^ 1 3+ 1 2x^ 2 3)^18. Then ( n m)^ 1 3 is:

Given expansion is ${(\frac{1}{3}{x}^{\frac{1}{3}}+\frac{1}{2{x}^{\frac{2}{3}}})}^{18}$(31​x31​+2x32​1​)18.

$$\Rightarrow {T}_{7}=C618{(\frac{1}{3}{x}^{\frac{1}{3}})}^{12}{(\frac{1}{2{x}^{\frac{2}{3}}})}^{6}$$⇒T7​=C618(31​x31​)12(2x32​1​)6

$$\Rightarrow m=C618{(\frac{1}{3})}^{12}{(\frac{1}{2})}^{6}$$⇒m=C618(31​)12(21​)6

$$\Rightarrow {T}_{13}=C1218{(\frac{1}{3}{x}^{\frac{1}{3}})}^{6}{(\frac{1}{2{x}^{\frac{2}{3}}})}^{12}$$⇒T13​=C1218(31​x31​)6(2x32​1​)12

$$\Rightarrow n=C1218{(\frac{1}{3})}^{6}{(\frac{1}{2})}^{12}$$⇒n=C1218(31​)6(21​)12

$$\Rightarrow \frac{m}{n}=\frac{C618{(\frac{1}{3})}^{12}{(\frac{1}{2})}^{6}}{C1218{(\frac{1}{3})}^{6}{(\frac{1}{2})}^{12}}$$⇒nm​=C1218(31​)6(21​)12C618(31​)12(21​)6​

$$\Rightarrow \frac{m}{n}=\frac{{(\frac{1}{3})}^{6}}{{(\frac{1}{2})}^{6}}$$⇒nm​=(21​)6(31​)6​

$$\Rightarrow \frac{m}{n}={(\frac{2}{3})}^{6}$$⇒nm​=(32​)6

$$\Rightarrow {(\frac{m}{n})}^{\frac{1}{3}}=\frac{4}{9}$$⇒(nm​)31​=94​

$$\Rightarrow {(\frac{n}{m})}^{\frac{1}{3}}=\frac{9}{4}$$⇒(mn​)31​=49​