Question

Let $\alpha>0$, be the smallest number such that the expansion of $\left(x^{\frac{2}{3}}+\frac{2}{x^3}\right)^{30}$ has a term $\beta x^{-a}, \beta \in N$.
Then \(α\) is equal to _________. 

Hint:

When dealing with binomial expansions, identify the general term and solve for the power of the variable to find the required term with the desired exponent.

Answer 1

Correct Answer: 2

The correct answer is 2.
$T_{r+1}={}^{30}{C}_r{\left(x^{2/3}\right)}^{30-r}{\left(\frac{2}{x^3}\right)}^r$
$={}^{30}{C}_r\cdot 2^r\cdot x^{\frac{60-11r}{3}}$
$\frac{60-11r}{3}<0\Rightarrow 11r>60\Rightarrow r>\frac{60}{11}\Rightarrow r=6$
$T_7={}^{30}{C}_6\cdot 2^6{x}^{-2}$
We have also observed $\beta ={}^{30}{C}_6(2{)}^6$ is a natural number.
$\therefore \alpha =2$

Answer 2

Step 1: General Term of the Expansion

The general term of the given expansion is:

\[ T_{r+1} = \binom{30}{r} \left( \frac{3}{x^3} \right)^{30-r} \left( \frac{2}{x} \right)^r. \]

We need to express this in the form \( \beta x^{-\alpha} \). The exponent of \( x \) in the general term is calculated as:

\[ \text{Power of } x = 3(30 - r) + (-r) = 90 - 4r. \]

Step 2: Setting the Power of \( x \)

For the term to be in the form \( x^{-\alpha} \), we set:

\[ 90 - 4r = -\alpha. \]

Step 3: Finding the Smallest Value of \( \alpha \)

Since \( \alpha > 0 \), we solve for \( r \):

\[ 60 - 11r > 60 \quad \Rightarrow \quad r = 6. \]

Step 4: Substituting \( r = 6 \) into the Expansion

We substitute \( r = 6 \) in the general term:

\[ T_7 = \binom{30}{6} 2^6 x^{-2}. \]

Step 5: Verifying \( \beta \)

We observe that \( \beta = \binom{30}{6} \times 2^6 \) is a natural number.

Step 6: Final Value of \( \alpha \)

From the exponent in \( x^{-2} \), we conclude:

\[ \alpha = 2. \]