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 ___

Hint:

When solving binomial expansion problems, focus on the general term and analyze the conditions for terms independent of a variable carefully.

Answer 1

Correct Answer: 1

The general term \(T_{r+1}\) in the binomial expansion is given by:
\[T_{r+1} = \binom{n}{r} (x)^r (y)^{n-r}.\]
In our case, we have:
\[T_{r+1} = \binom{22}{r} \left(x^{\frac{2}{3}}\right)^r (a)^{22 - 3r}.\]
Simplify:
\[T_{r+1} = \binom{22}{r} x^{\frac{2r}{3}} (a)^{22 - 3r}.\]
For the term independent of \(x\), the exponent of \(x\) must be zero. So:
\[\frac{2r}{3} = 0 \implies 44 - 11r = 0.\]
Solve for \(r\):
\[11r = 44 \implies r = 4.\]
Now, we substitute \(r = 4\) into the term \(T_{r+1}\):
\[T_5 = \binom{22}{4} a^4 = 7315.\]
Expand the binomial coefficient:
\[\binom{22}{4} = \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} = 7315.\]
Thus:
\[7315 a^4 = 7315 \implies a^4 = 1.\]
Solve for \(a\):
\[a = 1.\]
Conclusion
Thus, the value of \(a\) is \(1\).
Final Answer: The final answer is \(1\).