Question

If the seventh term from the beginning and end in the binomial expansion of ${(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}})}^n$ are equal, find n.

Answer 1

Given expansions is ${(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}})}^n$

$\therefore T_7=T_{6+1}=n_{C_6}{\left(\sqrt[3]{32}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{33}}\right)}^6$ ...(i)

Since, $T_7$ from end is same as the $T_7$ from beginning of ${(\frac{1}{\sqrt[3]{33}}+\sqrt[3]{32})}^2$

Then,

$T_7=n_{C_6}{\left(\frac{1}{\sqrt[3]{33}}\right)}^{n-6}{\left(\sqrt[3]{32}\right)}^6$ ...(ii)

Given that, $n_{C_6}\left(2\right)\frac{n-6}{3}\left(3\right)\frac{-6}{2}=n_{C_6}\left(3\right)\frac{(n-6)}{3}\left(2\right)\frac{6}{3}$

$\Rightarrow \left(2\right)\frac{n-12}{3}={\left(\frac{1}{3^{\frac{1}{3}}}\right)}^{n-12}$

which is true, when $\frac{n-12}{3}=0$

$\Rightarrow n-12=0\Rightarrow n=12$