Let f(x) = (2^{x+2} + 16) / (2^{2x+1} + 2^{x+4} + 32). Then the value of 8(f(1/15) + f(2/15) + ... +

Q: Wait, why do we simplify the function first? Can't we just plug in the values?

Great question! While you *could* theoretically plug in the values, you are dealing with fifty-nine terms involving powers of two. That would take hours and is a guaranteed path to a calculation error. In JEE Advanced, whenever you see a complex function in a summation, the function is almost always designed to simplify into something elegant. Always look for that simplification first!

Q: How did we know to check $f(4-x)$? Is there a rule for this?

Yes, there is! It's a powerful technique for summations. When you see a sum $\sum f(x_i)$ where the inputs $x_i$ have a symmetric property (like $x_1 + x_n = C$), it is a massive hint to check the behavior of $f(x) + f(C-x)$. If that sum is a constant, the entire series collapses into a simple multiplication problem.

Q: What happens to the middle term? Why did we add it separately?

Excellent observation. When you pair terms from the ends (first with last, second with second-to-last), you are essentially dividing the total number of terms by two. If the total number of terms is odd, there will always be one term left over in the middle that doesn't have a partner. In this case, with fifty-nine terms, we have twenty-nine pairs and one middle term. We must calculate that middle term's value individually and add it to the sum of the pairs.

Question

Let $f(x)=\frac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}.$ Then the value of $8(f(\frac{1}{15})+f(\frac{2}{15})+\dots+f(\frac{59}{15}))$ is equal to

Options:

(A) 92

(B) 118

(C) 102

(D) 108

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. When you first looked at this problem, I imagine your heart skipped a beat. A function

$$f(x) = \frac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}$$

and then a summation of fifty-nine terms? It looks like a nightmare designed to consume your entire exam time.

But here is the secret of the JEE Advanced: the more intimidating a problem looks, the more likely it is to have a beautiful, elegant shortcut hidden beneath the surface. Let's perform some algebraic surgery together.

### Simplifying the Function

We start by looking at the numerator: $2^{x+2} + 16$. This is just $2^x \cdot 2^2 + 16$, which is $4 \cdot 2^x + 16$. If we factor out a $4$, we get $4(2^x + 4)$.

Now, let's turn our attention to the denominator: $2^{2x+1} + 2^{x+4} + 32$. This can be rewritten as $2 \cdot (2^x)^2 + 16 \cdot 2^x + 32$.

If we factor out a $2$, we are left with $2((2^x)^2 + 8 \cdot 2^x + 16)$. Look closely at that quadratic expression inside the parenthesis; it is a perfect square, $(2^x + 4)^2$. So, our denominator is $2(2^x + 4)^2$.

Putting it all together, our function becomes:

$$f(x) = \frac{4(2^x + 4)}{2(2^x + 4)^2}$$

We can cancel the $2$ and one factor of $(2^x + 4)$. We are left with the beautifully simple:

$$f(x) = \frac{2}{2^x + 4}$$

### The Symmetry Insight

Now, look at the summation: $8 \sum_{k=1}^{59} f(\frac{k}{15})$. The inputs are $\frac{1}{15}, \frac{2}{15}, \dots, \frac{59}{15}$.

Notice the sum of the first and last inputs: $\frac{1}{15} + \frac{59}{15} = \frac{60}{15} = 4$. This is our golden ticket. Let's test the property of $f(x) + f(4-x)$.

Substituting $4-x$ into our simplified function, we get:

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

Using exponent rules, $2^{4-x} = \frac{16}{2^x}$. So, $f(4-x) = \frac{2}{\frac{16}{2^x} + 4}$. Multiplying the numerator and denominator by $2^x$, we get:

$$f(4-x) = \frac{2 \cdot 2^x}{16 + 4 \cdot 2^x} = \frac{2^x}{2(2^x + 4)}$$

Now, add them: $f(x) + f(4-x) = \frac{2}{2^x + 4} + \frac{2^x}{2(2^x + 4)}$. With a common denominator of $2(2^x + 4)$, the numerator becomes $4 + 2^x$.

This cancels perfectly with the denominator, leaving us with exactly $\frac{1}{2}$.

### The Summation Strategy

Because $f(x) + f(4-x) = \frac{1}{2}$, we can pair the terms in our sum. The first term $f(\frac{1}{15})$ pairs with the last term $f(\frac{59}{15})$ to give $\frac{1}{2}$.

With fifty-nine terms, we have twenty-nine such pairs, each summing to $\frac{1}{2}$. That gives us $29 	imes \frac{1}{2} = \frac{29}{2}$.

We have one term left in the middle: the $30^{th}$ term, where $x = \frac{30}{15} = 2$. We calculate:

$$f(2) = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}$$

Adding the pairs and the middle term:

$$S = \frac{29}{2} + \frac{1}{4} = \frac{58}{4} + \frac{1}{4} = \frac{59}{4}$$

Finally, the question asks for $8 	imes S$. So, $8 	imes \frac{59}{4} = 2 	imes 59 = 118$.

We have conquered the beast! Remember, in physics and math, the complexity is often just a mask. Your job is to look past it, find the symmetry, and let the algebra do the heavy lifting for you. The final answer is **118**.

Animated Solution for Mathematics - Functions: Let $f (x) = \frac{2 ^{x + 2} + 16}{2 ^{2 x + 1} + 2 ^{x + 4} + 32} .$ Then the value of $8 (f (\frac{1}{15}) + f (\frac{2}{15}) + \dots + f (\frac{59}{15}))$ is equal to

Select Answer:

Visualized Solution

Analyze the Function $f (x)$

Simplify the Numerator

Simplify the Denominator

The Simplified Function

Explore Symmetry

Simplify $f (4 - x)$

The Magic Property

Analyze the Series Pairs

Continuing the Pairing

Count the Pairs

The Middle Term

Calculate Total Sum $S$

Final Answer

The Sigma Insight: Classification of Functions

Analyzing the Setup

Simplifying the Function

The Symmetry Insight

The Summation Strategy

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