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 $\beta x^{-\alpha}, \beta \in \mathbb{N}$. Then $\alpha$ is equal to ___________.
Answer (integer)
2
Solution
$\mathrm{T}_{\mathrm{r}+1}={ }^{30} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2 / 3}\right)^{30-\mathrm{r}}\left(\frac{2}{\mathrm{x}^{3}}\right)^{\mathrm{r}}$
<br/><br/>$={ }^{30} \mathrm{C}_{\mathrm{r}} \cdot 2^{\mathrm{r}} \cdot \mathrm{x}^{\frac{60-11 \mathrm{r}}{3}}$
<br/><br/>$\frac{60-11 \mathrm{r}}{3}<0
$
<br/><br/>$\Rightarrow 11 \mathrm{r}>60 $
<br/><br/>$\Rightarrow \mathrm{r}>\frac{60}{11} $
<br/><br/>$\Rightarrow \mathrm{r}=6$
<br/><br/>$\mathrm{T}_{7}={ }^{30} \mathrm{C}_{6} \cdot 2^{6} \mathrm{x}^{-2}$
<br/><br/>We have also observed $\beta={ }^{30} \mathrm{C}_{6}(2)^{6}$ is a natural number.
<br/><br/>$\therefore \alpha=2$
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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