Find the third term of the expansion of $\left(z^2 + \dfrac{1}{z} \sqrt[3]{z}\right)^n$ if the sum of all the binomial coefficients is equal to $2048$.
Sum of all binomial coefficients is
$C^n_0 + C^n_1 + C^n_2 + \cdots + C^n_n = 2^n$
Then, as per question,
$2^n = 2048$
i.e. $\;$ $2^n = 2^{11}$ $\implies$ $n = 11$
$\therefore \;$ the problem is to find the third term in the expansion of
$\begin{aligned}
\left(z^2 + \dfrac{1}{z} \sqrt[3]{z}\right)^n & = \left(z^2 + z^{\frac{1}{3} - 1}\right)^{11} \\\\
& = \left(z^2 + z^{\frac{-2}{3}}\right)^{11}
\end{aligned}$
$\therefore \;$ The required third term is
$\begin{aligned}
T_3 = T_{2+1} & = C^{11}_{2} \; \left(z^2\right)^{11 - 2} \; \left(z^{\frac{-2}{3}}\right)^2 \\\\
& = \dfrac{11!}{2! \times \left(11 - 2\right)!} \times \left(z^2\right)^9 \times z^{\frac{-4}{3}} \\\\
& = \dfrac{11 \times 10 \times 9!}{2 \times 9!} \times z^{18 - \frac{4}{3}} \\\\
& = 55 z^{\frac{50}{3}}
\end{aligned}$