Question

Let $\alpha >0$, be the smallest number such that the expansion of ${\left(x^{\frac{2}{3}}+\frac{2}{x^3}\right)}^{30}$ has a term $\beta x^{-a},\beta \in N$. Then $\alpha$ is equal to

Answer 1

Answer: 2

$T_{r+1}={}^{30}{C}_r{\left(x^{2/3}\right)}^{30-r}{\left(\frac{2}{x^3}\right)}^r$
$={}^{30}{C}_r\cdot 2^r\cdot x^{\frac{60-11r}{3}}$
$\frac{60-11r}{3}<0\Rightarrow 11r>60\Rightarrow r>\frac{60}{11}\Rightarrow r=6$
$T_7={}^{30}{C}_6\cdot 2^6{x}^{-2}$
We have also observed $\beta ={}^{30}{C}_6(2{)}^6$ is a natural number.
$\therefore \alpha =2$