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In the binomial \(\left(2^{1 / 3}+3^{-1 / 3}\right)^{\mathrm{n}}\), if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is \(1 / 6\), then \(n\) equal to
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The correct answer is:
9
\(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} a^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{b}^{\mathrm{r}}\) where \(\mathrm{a}=2^{1 / 3}\) and \(\mathrm{b}=3^{-1 / 3}\)
\(\mathrm{T}_7\) from beginning \(={ }^n \mathrm{C}_6 a^{\mathrm{n}-6} \mathrm{~b}^6\) and \(\mathrm{T}_7\) from end \(={ }^{\mathrm{n}} \mathrm{C}_6 \quad \mathrm{~b}^{\mathrm{n}-6} \quad \mathrm{a}^6\)
\(\begin{aligned}
& \Rightarrow \frac{\mathrm{a}^{\mathrm{n}-12}}{\mathrm{~b}^{\mathrm{n}-12}}=\frac{1}{6} \Rightarrow 2^{\frac{\mathrm{n}-12}{3}} \cdot 3^{\frac{\mathrm{n}-12}{3}}=6^{-1} \\
& \Rightarrow \mathrm{n}-12=-3 \Rightarrow \mathrm{n}=9
\end{aligned}\)
\(\mathrm{T}_7\) from beginning \(={ }^n \mathrm{C}_6 a^{\mathrm{n}-6} \mathrm{~b}^6\) and \(\mathrm{T}_7\) from end \(={ }^{\mathrm{n}} \mathrm{C}_6 \quad \mathrm{~b}^{\mathrm{n}-6} \quad \mathrm{a}^6\)
\(\begin{aligned}
& \Rightarrow \frac{\mathrm{a}^{\mathrm{n}-12}}{\mathrm{~b}^{\mathrm{n}-12}}=\frac{1}{6} \Rightarrow 2^{\frac{\mathrm{n}-12}{3}} \cdot 3^{\frac{\mathrm{n}-12}{3}}=6^{-1} \\
& \Rightarrow \mathrm{n}-12=-3 \Rightarrow \mathrm{n}=9
\end{aligned}\)
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