Key Concepts and Formulas

• Derivatives for Optimization: To find the maximum or minimum of a continuous function, we find critical points by setting the first derivative equal to zero.
• Quotient Rule: If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.
• First Derivative Test: If $f'(x)$ changes from positive to negative at $x=c$, then $f(x)$ has a local maximum at $x=c$.

Step-by-Step Solution

Step 1: Define the Continuous Function

We are given the sequence $a_n = \frac{n^3}{n^4 + 147}$ for $n = 1, 2, 3, \ldots$. To use calculus, we define a continuous function $f(x)$ such that $f(x) = \frac{x^3}{x^4 + 147}$ for $x \geq 1$. This allows us to use derivatives to find the maximum value.

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

Step 2: Calculate the First Derivative

To find the critical points, we need to find the first derivative $f'(x)$. We use the quotient rule where $u(x) = x^3$ and $v(x) = x^4 + 147$. Thus, $u'(x) = 3x^2$ and $v'(x) = 4x^3$.

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

Simplifying the numerator:

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

Factoring out $x^2$ from the numerator:

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

Step 3: Find the Critical Points

We set $f'(x) = 0$ to find the critical points. Since the denominator is always positive for real $x$, we only need to consider when the numerator is zero.

$x^2(441 - x^4) = 0$

This gives us $x^2 = 0 \Rightarrow x = 0$ or $441 - x^4 = 0 \Rightarrow x^4 = 441$. Since $x \geq 1$, $x=0$ is not relevant. Solving $x^4 = 441$:

$x = \sqrt[4]{441} = \sqrt{\sqrt{441}} = \sqrt{21}$

So, the critical point is $x = \sqrt{21}$.

Step 4: Apply the First Derivative Test

We analyze the sign of $f'(x)$ around $x = \sqrt{21}$. Since $x^2$ and $(x^4 + 147)^2$ are always positive, the sign of $f'(x)$ depends on the sign of $(441 - x^4)$.

• If $x < \sqrt{21}$, then $x^4 < 441$, so $441 - x^4 > 0$, and $f'(x) > 0$.
• If $x > \sqrt{21}$, then $x^4 > 441$, so $441 - x^4 < 0$, and $f'(x) < 0$.

Since $f'(x)$ changes from positive to negative at $x = \sqrt{21}$, there is a local maximum at $x = \sqrt{21}$.

Step 5: Identify Candidate Integer Values for n

Since $f(x)$ has a maximum at $x = \sqrt{21}$, we need to check integer values of $n$ closest to $\sqrt{21}$. We know that $4^2 = 16$ and $5^2 = 25$, so $4 < \sqrt{21} < 5$. Since $\sqrt{21} \approx 4.58$, the integer values to consider are $n = 4$ and $n = 5$.

Step 6: Evaluate the Sequence Terms at Candidate Integers

We evaluate $a_n$ for $n = 4$ and $n = 5$:

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

Since $a_5 > a_4$, the greatest term in the sequence occurs at $n = 5$.

Step 7: Determine the value of alpha

Since $a_{\alpha}$ is the greatest term in the sequence, and we found that the greatest term is $a_5$, then $\alpha = 5$.

Common Mistakes & Tips

• Remember to check integer values around the critical point since the sequence is only defined for integers.
• Double-check the derivative calculation, as errors there can lead to incorrect critical points.
• Be careful with inequalities when determining the sign of the derivative.

Summary

To find the greatest term in the sequence, we created a continuous function, found its maximum using derivatives, and then checked the integer values of $n$ around the maximum to determine the largest term. The greatest term is $a_5$, thus $\alpha = 5$.

The final answer is $\boxed{5}$.