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Let α>0\alpha>0α>0, be the smallest number such that the expansion of (x23+2x3)30\left(x^{\frac{2}{3}}+\frac{2}{x^{3}}\right)^{30}(x32+x32)30 has a term βx−α,β∈N\beta x^{-\alpha}, \beta \in \mathbb{N}βx−α,β∈N. Then α\alphaα is equal to ___________.
Tr+1= 30 Cr(x2 / 3)30−r(2x3)r\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}}Tr+1= 30 Cr(x2 / 3)30−r(x32)r
= 30 Cr ⋅ 2r ⋅ x60−11 r3={ }^{30} \mathrm{C}_{\mathrm{r}} \cdot 2^{\mathrm{r}} \cdot \mathrm{x}^{\frac{60-11 \mathrm{r}}{3}}= 30 Cr ⋅ 2r ⋅ x360−11 r
60−11 r3<0 \frac{60-11 \mathrm{r}}{3}<0 360−11 r<0
⇒ 11 r>60 \Rightarrow 11 \mathrm{r}>60 ⇒ 11 r>60
⇒ r>6011 \Rightarrow \mathrm{r}>\frac{60}{11} ⇒ r>1160
⇒ r=6\Rightarrow \mathrm{r}=6⇒ r=6
T7= 30 C6 ⋅ 26 x−2\mathrm{T}_{7}={ }^{30} \mathrm{C}_{6} \cdot 2^{6} \mathrm{x}^{-2}T7= 30 C6 ⋅ 26 x−2
We have also observed β= 30 C6(2)6\beta={ }^{30} \mathrm{C}_{6}(2)^{6}β= 30 C6(2)6 is a natural number.
∴ α=2\therefore \alpha=2∴ α=2
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