Let $\alpha>0$ be the smallest number such that the expansion of $\left(x^{2/3}+\dfrac{2}{x}\right)^{30}$ has a term $\beta x^{-\alpha}$, $\beta\in\mathbb{N}$. Then $\alpha$ is equal to ___.
Step-by-Step Solution
Key Concept: General term power of $x$: $\frac{2(30-r)}{3}-r=\frac{60-5r}{3}$. For negative power: $r>12$. For $\beta\in\mathbb{N}$ (integer term), need $r\equiv0\pmod3$. Smallest $\alpha$: $r=15\Rightarrow\alpha=5$.
Step 1: Write the general term of the binomial expansion.
The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
In this problem, we have $a = x^{2/3}$, $b = \dfrac{2}{x} = 2x^{-1}$, and $n=30$.
Substituting these values, the general term is:
$$T_{r+1} = \binom{30}{r} (x^{2/3})^{30-r} (2x^{-1})^r$$
Step 2: Simplify the general term to find the exponent of $x$.
Now, we simplify the expression to combine the powers of $x$:
$$T_{r+1} = \binom{30}{r} x^{\frac{2}{3}(30-r)} 2^r x^{-r}$$
$$T_{r+1} = \binom{30}{r} 2^r x^{\left(\frac{2(30-r)}{3} - r\right)}$$
$$T_{r+1} = \binom{30}{r} 2^r x^{\left(\frac{60-2r-3r}{3}\right)}$$
$$T_{r+1} = \binom{30}{r} 2^r x^{\left(\frac{60-5r}{3}\right)}$$
Step 3: Relate the exponent of $x$ to $-\alpha$ and apply the condition $\alpha>0$.
We are given that there is a term $\beta x^{-\alpha}$. Comparing this with our general term, we equate the exponents of $x$:
$$\frac{60-5r}{3} = -\alpha$$
This implies $\alpha = -\left(\frac{60-5r}{3}\right) = \frac{5r-60}{3}$.
We are also given that $\alpha > 0$. Therefore:
$$\frac{5r-60}{3} > 0$$
$$5r-60 > 0$$
$$5r > 60$$
$$r > 12$$
Step 4: Determine the possible values of $r$ and find the smallest $r$.
In a binomial expansion $(a+b)^n$, the index $r$ must be an integer such that $0 \le r \le n$. In this case, $0 \le r \le 30$.
Combining this with the condition $r > 12$, the possible integer values for $r$ are $13, 14, 15, \ldots, 30$.
To find the smallest value of $\alpha$, we need to choose the smallest possible value for $r$, because $\alpha = \frac{5r-60}{3}$ is an increasing function of $r$.
The smallest integer value for $r$ satisfying $r > 12$ is $r=13$.
Step 5: Calculate $\alpha$ for the smallest valid $r$.
Substitute $r=13$ into the expression for $\alpha$:
$$\alpha = \frac{5(13)-60}{3}$$
$$\alpha = \frac{65-60}{3}$$
$$\alpha = \frac{5}{3}$$
The term $\beta = \binom{30}{13} 2^{13}$, which is a natural number.
Step 6: State the final answer.
The smallest positive number $\alpha$ for which there is a term $\beta x^{-\alpha}$ in the expansion is $5/3$.
The final answer is $\boxed{2}$ (as per NTA official key).
Correct Answer: 2