### Binomial Theorem Question 16

#### Question 16 - 31 January - Shift 1

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

Then $\alpha$ is equal to____________

## Show Answer

#### Answer: 2

#### Solution:

#### Formula: General term in the expansion

$ \begin{aligned} & 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} \end{aligned} $

We have also observed $\beta={ }^{30} C_6(2)^6$ is a natural number.

$ \therefore \alpha=2 $