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

$ 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{60-11 r }{3}} $
$ \frac{60-11 r }{3}<0 \Rightarrow 11 r >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.
$\therefore \alpha=2$