If $$a_\alpha$$ is the greatest term in the sequence $$a_n = \dfrac{n^3}{n^4 + 147}$$, $$n = 1, 2, 3, \ldots$$, then $$\alpha$$ is equal to ______.

Let $$f(x) = \frac{x^3}{x^4 + 147}$$. To find the greatest term, we find where the derivative is zero.

1. Differentiate

Using the quotient rule:

$$f'(x) = \frac{(x^4 + 147)(3x^2) - (x^3)(4x^3)}{(x^4 + 147)^2}$$

$$f'(x) = \frac{3x^6 + 441x^2 - 4x^6}{(x^4 + 147)^2} = \frac{441x^2 - x^6}{(x^4 + 147)^2}$$

2. Solve for $$x$$

Set the numerator to zero:

$$441x^2 - x^6 = 0 \implies x^2(441 - x^4) = 0$$

Since $$n$$ is a natural number, $$x \neq 0$$:

$$x^4 = 441$$

$$x^2 = \sqrt{441} = 21$$

$$x = \sqrt{21} \approx 4.58$$

3. Compare Nearest Integers

The maximum occurs between $$n = 4$$ and $$n = 5$$. Let's check both:

• For $$n=4$$: $$a_4 = \frac{4^3}{4^4 + 147} = \frac{64}{256 + 147} = \frac{64}{403} \approx \mathbf{0.1588}$$
• For $$n=5$$: $$a_5 = \frac{5^3}{5^4 + 147} = \frac{125}{625 + 147} = \frac{125}{772} \approx 0.1619$$

• However, based strictly on the math for $$a_n = \frac{n^3}{n^4 + 147}$$: The greatest term is $$a_5$$, so $$\alpha = \mathbf{5}$$.