Question

Let $P$ be the $7^{\text {th }}$ term from the beginning and $Q$ be the $7^{\text {th }}$ term from the end in the expansion of $\left(\sqrt[3]{3}+\frac{1}{\sqrt[3]{4}}\right)^{ n }$ where $n \in N$. If $12 P = Q$, then find the value of $n$.

Answer 1

$P={ }^n C_6\left(3^{\frac{1}{3}}\right)^{n-6} \cdot\left(4^{\frac{-1}{3}}\right)^6$
$Q ={ }^{ n } C _{ n -6}\left(3^{\frac{1}{3}}\right)^6 \cdot\left(4^{\frac{-1}{3}}\right)^{ n -6}$
$\therefore \frac{ Q }{ P }=12 \Rightarrow(12)^{\frac{ n -6}{3}}=(12)^1 \Rightarrow \frac{ n -6}{3}=1 \Rightarrow n =9$