Key Concept: The Binomial Theorem

For any binomial expression of the form $(a+b)^n$, the general term (or the $(r+1)^{th}$ term) in its expansion is given by the Binomial Theorem: $T_{r+1} = {n}C_{r} \, a^{n-r} \, b^{r}$ where $n$ is a non-negative integer, and ${n}C_{r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient. This formula allows us to find any specific term in an expansion without writing out the entire series.

Step-by-Step Solution

1. Identify $a$, $b$, and $n$ from the given expansion: The given expression is $\left(x^{\frac{2}{3}}+\frac{2}{x^{3}}\right)^{30}$. Comparing this with $(a+b)^n$:

• $a = x^{\frac{2}{3}}$
• $b = \frac{2}{x^{3}}$
• $n = 30$

2. Write down the general term $T_{r+1}$: Using the binomial theorem formula, the $(r+1)^{th}$ term is: $T_{r+1} = {30}C_{r} \left(x^{\frac{2}{3}}\right)^{30-r} \left(\frac{2}{x^{3}}\right)^{r}$ Explanation: This step applies the Binomial Theorem directly to our given expression to get a generic representation of any term in its expansion. The variable $r$ will be an integer from $0$ to $30$.

3. Simplify the general term, combining powers of $x$: $T_{r+1} = {30}C_{r} \, x^{\frac{2}{3}(30-r)} \, \frac{2^{r}}{(x^{3})^{r}}$ $T_{r+1} = {30}C_{r} \, x^{\frac{60-2r}{3}} \, 2^{r} \, x^{-3r}$ Now, combine the terms with $x$ using the rule $x^m \cdot x^p = x^{m+p}$: $T_{r+1} = {30}C_{r} \, 2^{r} \, x^{\frac{60-2r}{3} - 3r}$ To combine the exponents, find a common denominator: $T_{r+1} = {30}C_{r} \, 2^{r} \, x^{\frac{60-2r - 9r}{3}}$ $T_{r+1} = {30}C_{r} \, 2^{r} \, x^{\frac{60-11r}{3}}$ Explanation: This simplification is crucial to isolate the exponent of $x$. We use exponent rules $(x^m)^p = x^{mp}$ and $x^{-p} = \frac{1}{x^p}$ to bring all $x$ terms to the numerator and then combine them.

4. Determine the condition for the term to be of the form $\beta x^{-\alpha}$ with $\alpha > 0$: The problem states that there is a term $\beta x^{-\alpha}$, where $\alpha > 0$. This means the exponent of $x$ in our general term must be a negative value. So, we must have: $\frac{60-11r}{3} < 0$ Since $3 > 0$, we can multiply both sides by $3$ without changing the inequality direction: $60 - 11r < 0$ $60 < 11r$ $r > \frac{60}{11}$ Explanation: The requirement $\alpha > 0$ for a term $\beta x^{-\alpha}$ implies the exponent of $x$ must be negative. We set up an inequality to find the values of $r$ that satisfy this condition.

5. Find the smallest integer value of $r$ that satisfies the condition: We know that $r$ must be an integer, representing the power of the second term in the binomial expansion. Calculating the value of $\frac{60}{11}$: $\frac{60}{11} \approx 5.45$ So, $r > 5.45$. Since $r$ must be an integer, the smallest integer value of $r$ that satisfies this condition is $r=6$. Explanation: The problem asks for the smallest number $\alpha > 0$. The exponent of $x$ is $\frac{60-11r}{3}$. For this exponent to be negative, $r$ must be greater than $60/11$. As $r$ increases, the term $11r$ increases, making the numerator $60-11r$ more negative. Therefore, to get the smallest positive $\alpha$ (which is the magnitude of the negative exponent), we need the largest negative exponent closest to zero. This corresponds to the smallest possible integer value of $r$ that makes the exponent negative, which is $r=6$.

6. Substitute $r=6$ back into the general term to find $\alpha$ and $\beta$: With $r=6$, the term is $T_{6+1} = T_7$: $T_7 = {30}C_{6} \, 2^{6} \, x^{\frac{60-11(6)}{3}}$ $T_7 = {30}C_{6} \, 2^{6} \, x^{\frac{60-66}{3}}$ $T_7 = {30}C_{6} \, 2^{6} \, x^{\frac{-6}{3}}$ $T_7 = {30}C_{6} \, 2^{6} \, x^{-2}$ Comparing this term with $\beta x^{-\alpha}$:

• The exponent of $x$ is $-2$, so $-\alpha = -2$, which means $\alpha = 2$.
• The coefficient is ${30}C_{6} \cdot 2^{6}$, so $\beta = {30}C_{6} \cdot 2^{6}$.

7. Verify that $\beta$ is a natural number:

• ${30}C_{6}$ is a binomial coefficient, which is always a positive integer.
• $2^{6} = 64$, which is a positive integer.
• The product of two positive integers is a positive integer. Therefore, $\beta = {30}C_{6} \cdot 2^{6}$ is indeed a natural number ($\mathbb{N}$), satisfying the condition given in the problem.

Conclusion: Based on our calculations, the smallest positive number $\alpha$ for which a term $\beta x^{-\alpha}$ exists in the expansion is $\alpha = 2$.

Key Takeaway & Common Mistakes:

• Understanding "smallest $\alpha > 0$": This means we are looking for the least negative integer exponent for $x$ (closest to zero) in the expansion. If the problem asked for the largest $\alpha$, we would look for the largest $r$ that makes the exponent negative.
• Careful with Exponent Rules: Mistakes often occur when simplifying powers of $x$, especially with fractions and negative exponents.
• Checking Constraints: Always verify that the calculated $\beta$ satisfies any given conditions (e.g., $\beta \in \mathbb{N}$).
• Integer values for r: Remember that $r$ in the binomial expansion must be a non-negative integer.

The final answer is $\boxed{2}$.