Animated Solution for Mathematics - Functions: If $f(x)=\frac{2^{2x}}{2^{2x}+2},x\in \mathbb{R}$, then $f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\cdots +f\left(\frac{2022}{2023}\right)$ is equal to

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a daunting mountain of arithmetic.

You see a series: $S=f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\cdots +f\left(\frac{2022}{2023}\right)$.

Your instinct might be to panic, thinking you need to calculate each term individually. But in the world of JEE Advanced, whenever you see a long, intimidating sum, there is almost always a hidden symmetry waiting to be uncovered. Let us embark on this journey together.

The Simplification

Before we touch the series, we must understand the function $f(x)=\frac{2^{2x}}{2^{2x}+2}$. That $2^{2x}$ looks a bit clunky, doesn't it?

We know that $2^{2x}=(2^2{)}^x=4^x$. Substituting this back into our function, we get:

This is much cleaner. It is the same function, just dressed in a more elegant outfit. Always look for ways to simplify your expressions before diving into the heavy lifting.

The Symmetry Hunt

Now, let us look at the inputs of our series: $\frac{1}{2023},\frac{2}{2023},\dots ,\frac{2022}{2023}$.

Notice something? The first input is $\frac{1}{2023}$ and the last is $\frac{2022}{2023}$. If we add them, we get:

This is the 'Aha!' moment. This suggests that the function might have a special property when evaluated at $x$ and $1-x$. Let us test this hypothesis by calculating $f(1-x)$.

The Algebraic Dance

Substituting $1-x$ into our simplified function, we get:

Now, let us use the property $a^{m-n}=\frac{a^m}{a^n}$ to expand $4^{1-x}$. This gives us:

To clear this complex fraction, we multiply the numerator and the denominator by $4^x$. The expression transforms into:

Finally, dividing both the numerator and the denominator by $2$, we arrive at:

The Revelation

Now, for the magic. Let us add $f(x)$ and $f(1-x)$ together:

Since the denominators are identical, we simply add the numerators:

This is the core of the problem! We have proven that $f(x)+f(1-x)=1$. This means that any two terms whose inputs add up to $1$ will sum to $1$.

The Grand Finale

We have $2022$ terms in our series. We can pair the first term with the last, the second with the second-to-last, and so on.

Since each pair sums to $1$, and we have $\frac{2022}{2}=1011$ pairs, the total sum is simply:

We have conquered the mountain, not by brute force, but by understanding the elegant structure beneath. Keep this property in your toolkit—it is a powerful weapon for your JEE journey. The final answer is 1011.