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 (integer)
1
Solution
Given expansion $\left(x^{\frac{2}{3}}+\frac{\alpha}{x^3}\right)^{22}$
<br/><br/>$$
T_{r+1}={ }^{22} C_r\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r
$$
<br/><br/>For constant term
<br/><br/>$$
\begin{aligned}
& \frac{44-2 r}{3}-3 r=0 \\\\
& \Rightarrow r=4
\end{aligned}
$$
<br/><br/>Now ${ }^{22} \mathrm{C}_4 \alpha^4=7315$
<br/><br/>$$
\begin{aligned}
& \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} \alpha^4=7315 \\\\
& \therefore \alpha^4=1 \\\\
& \therefore |\alpha|=1
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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