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} & \mathrm{T}{\mathrm{r}+1}={ }^{30} \mathrm{C}{\mathrm{r}}\left(\mathrm{x}^{2 / 3}\right)^{30-\mathrm{r}}\left(\frac{2}{\mathrm{x}^3}\right)^{\mathrm{r}} \\ & ={ }^{30} \mathrm{C}_{\mathrm{r}} \cdot 2^{\mathrm{r}} \cdot \mathrm{x}^{\frac{60-11 \mathrm{r}}{3}} \\ & \frac{60-11 \mathrm{r}}{3}<0 \Rightarrow 11 \mathrm{r}>60 \\ & \Rightarrow \mathrm{r}>\frac{60}{11} \Rightarrow \mathrm{r}=6 \\ & \mathrm{~T}_7={ }^{30} \mathrm{C}_6 \cdot 2^6 \mathrm{x}^{-2} \end{aligned} $
We have also observed $\beta={ }^{30} C_6(2)^6$ is a natural number.
$ \therefore \alpha=2 $
