Question

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

Answer 1

Answer: 1

$T_{r+1}={}^{22}{C}_r\cdot {\left(x^{\frac{2}{3}}\right)}^{22-r}\cdot (\alpha {)}^r,x^{-3r}$
$={}^{22}{C}_r\cdot x^{\frac{44}{3}-\frac{2r}{3}-3r}(\alpha {)}^r$
$\frac{44}{3}=\frac{11r}{3}$
$r=4$
${}^{22}{C}_4\cdot {\alpha }^4=7315$
$\frac{22\times 21\times 20\times 19}{24}\cdot {\alpha }^4=7315$
$\alpha =1$