Binomial Theorem Question 22
Question 22 - 01 February - Shift 2
Sum of the squares of all possible values of $x=4$ If the term without $x$ in the expansion of $(x^{\frac{2}{3}}+\frac{\alpha}{x^{3}})^{22}$ is 7315 , then $|\alpha|$ is equal to________
Show Answer
Answer: (1)
Solution:
Formula: General term in the expansion
$T _{r+1}={ }^{22} C_r \cdot(x^{\frac{2}{3}})^{22-r} \cdot(\alpha)^{r}, x^{-3 r}$
$={ }^{22} C_r \cdot x^{\frac{44}{3}-\frac{2 r}{3}-3 r}(\alpha)^{r}$
$\frac{44}{3}=\frac{11 r}{3}$
$r=4$
${ }^{22} C_4 \cdot \alpha^{4}=7315$
$22 \times 21 \times 20 \times 19$
24
$\alpha^{4}=7315$
$\alpha=1$
