If a_α is the greatest term in the sequence a_n = n³/(n⁴ + 147), n = 1, 2, 3, ..., then α is equal t

Q: Wait, why can I just replace $n$ with $x$ and use calculus? The sequence is discrete, not continuous!

That is a brilliant observation! You are right that the sequence is discrete. However, by defining a continuous function $f(x)$ that matches the sequence at every integer point, we gain the power of calculus to find the 'neighborhood' of the maximum. Once we find the peak of the continuous curve, we know the discrete maximum must be one of the integers flanking that peak.

Q: Why did we set the derivative to zero? Could it be a minimum instead of a maximum?

Great catch! Setting the derivative to zero only identifies critical points. To be absolutely sure it is a maximum, you could check the second derivative or simply compare the values of the terms around the critical point. In this problem, comparing $a_4$ and $a_5$ is the most robust way to confirm which one is truly the greatest.

Q: Is there a shortcut? Calculating $a_4$ and $a_5$ with those large denominators feels tedious.

In the heat of the JEE Advanced exam, precision is your best friend. While you could try to estimate, calculating the exact values is the safest path. Think of it as a final check to ensure your calculus didn't lead you astray. It only takes a few seconds to verify!

Q: What if the critical point was exactly an integer? Would that be the answer?

Yes, absolutely! If the critical point $x$ had turned out to be an integer (like $x=5$), then $a_5$ would be the global maximum of the continuous function, and consequently, the greatest term of the sequence. We only check neighbors when the peak falls between two integers.

Q: Why did we ignore the $x=0$ solution from the derivative?

Look at the domain of our sequence: $n = 1, 2, 3, \dots$. Since $n$ starts at 1, $x=0$ is outside our region of interest. Always keep an eye on your domain constraints—they are the boundaries of your mathematical playground!

Question

If $a_\alpha$ is the greatest term in the sequence $a_n = \frac{n^3}{n^4 + 147}, n = 1, 2, 3, \dots$, then $\alpha$ is equal to ——————

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the base of a mountain range. Each step you take represents an integer $n=1, 2, 3, \dots$.

Your goal is to find the highest point you can reach, the 'greatest term' of the sequence defined by:

$$a_n = \frac{n^3}{n^4 + 147}$$

This isn't just a math problem; it is a journey of discovery. We are looking for the summit of a function that governs the behavior of these numbers.

### Bridging the Discrete and the Continuous

When we look at a sequence, it can feel like a series of disconnected dots. To understand the 'shape' of these dots, we perform a powerful transformation: we treat the sequence as a continuous function, $f(x) = \frac{x^3}{x^4 + 147}$, where $x \ge 1$.

By doing this, we move from the world of discrete arithmetic into the elegant, flowing world of calculus. We are no longer just guessing; we are mapping the terrain.

### The Power of the Derivative

To find the peak of any smooth curve, we must look for the point where the slope is zero. This is where the magic of the **Quotient Rule** comes into play.

We define our function as $f(x) = \frac{u}{v}$, where $u = x^3$ and $v = x^4 + 147$. The rule tells us that the derivative is given by:

$$f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2}$$

As we substitute our values, we get:

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

When we expand the numerator, the terms simplify beautifully. We are left with:

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

### Finding the Critical Point

Setting $f'(x) = 0$ reveals the critical point where the function stops rising and starts falling. Since $x^2$ cannot be zero in our domain, we focus on the term $(441 - x^4) = 0$.

This leads us to $x^4 = 441$, which means $x^2 = 21$. Calculating the square root, we find $x = \sqrt{21} \approx 4.58$.

This is the 'theoretical peak' of our continuous function. Since $4.58$ lies between $4$ and $5$, the highest point of our sequence must be at one of these two integers.

### The Final Verification

Now, we test our candidates by calculating the values for $n=4$ and $n=5$:

For $n=4$:

$$a_4 = \frac{4^3}{4^4 + 147} = \frac{64}{256 + 147} = \frac{64}{403} \approx 0.1588$$

For $n=5$:

$$a_5 = \frac{5^3}{5^4 + 147} = \frac{125}{625 + 147} = \frac{125}{772} \approx 0.1619$$

Comparing these two, we see that $a_5 > a_4$. The peak of our mountain is indeed at $n=5$.

### Conclusion: The Elegance of the Result

By using the tools of calculus, we transformed a daunting sequence into a clear, visualizable path. We didn't just find the answer; we understood the landscape.

The greatest term is **$\alpha = 5$**. Always remember, in JEE Advanced, the math is not just about calculation—it is about the narrative of the function.

Animated Solution for Mathematics - Differentiation: If $a_{α}$ is the greatest term in the sequence $a_{n} = \frac{n ^{3}}{n ^{4} + 147}, n = 1, 2, 3, \dots$ , then $α$ is equal to ——————

Enter Numerical Value:

Visualized Solution

Define the Continuous Function $f (x)$

Quotient Rule for Differentiation

Apply Quotient Rule

Calculate Derivatives

Simplify the Numerator

Find Critical Points

Solve for $x$

Identify Neighboring Integers

Calculate $a_{4}$

Calculate $a_{5}$

Conclusion: The Greatest Term

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Bridging the Discrete and the Continuous

The Power of the Derivative

Finding the Critical Point

The Final Verification

Conclusion

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If aα​ is the greatest term in the sequence an​=n4+147n3​,n=1,2,3,…, then α is equal to ——————