Let 0 be the smallest number such that the expansion of (x^2/3+2x^3)^30 has a term x^-, N…

Question

Let $\alpha>0$ be the smallest number such that the expansion of $\left(x^{2/3}+\dfrac{2}{x^3}\right)^{30}$ has a term $\beta x^{-\alpha}$, $\beta\in N$. Then $\alpha$ is equal to \_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

$$T_{r+1}=\binom{30}{r}(x^{2/3})^{30-r}\left(\frac{2}{x^3}\right)^r=\binom{30}{r}2^r\,x^{\frac{60-2r}{3}-3r}=\binom{30}{r}2^r\,x^{\frac{60-11r}{3}}$$

For a negative power ($x^{-\alpha}$, $\alpha>0$): $\dfrac{60-11r}{3}<0\Rightarrow r>\dfrac{60}{11}\Rightarrow r\ge6$.

At $r=6$: power $=\dfrac{60-66}{3}=-2$, giving $T_7=\binom{30}{6}2^6x^{-2}$, with $\beta=\binom{30}{6}2^6\in N$.

$$\alpha=2$$

Solutions · 1