If the term without x in the expansion of (x^2/3+x^3)^22 is 7315, then || is equal to ____…

Question

If the term without $x$ in the expansion of $\left(x^{2/3}+\dfrac{\alpha}{x^3}\right)^{22}$ is 7315, then $|\alpha|$ is equal to \_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

$$T_{r+1}=\binom{22}{r}(x^{2/3})^{22-r}(\alpha x^{-3})^r=\binom{22}{r}\alpha^r\,x^{\frac{44-2r}{3}-3r}=\binom{22}{r}\alpha^r\,x^{\frac{44-11r}{3}}$$

Setting exponent $=0$: $44=11r\Rightarrow r=4$.

$$\binom{22}{4}\alpha^4=7315\quad\Rightarrow\quad\frac{22\times21\times20\times19}{24}\alpha^4=7315\quad\Rightarrow\quad7315\,\alpha^4=7315\quad\Rightarrow\quad\alpha^4=1$$

$$|\alpha|=1$$

Solutions · 1