Animated Solution for Mathematics - Functions: If $f(x)=\frac{2^x}{2^x+\sqrt{2}},x\in \mathbb{R},$ then ${\sum }_{k=1}^{81}f(\frac{k}{82})$ is equal to

Mastering Symmetry in Summation

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of calculation. You see a summation from $k=1$ to $81$ of a function $f(x)=\frac{2^x}{2^x+\sqrt{2}}$.

Your instinct might be to start plugging in values, to calculate $f(\frac{1}{82})$, then $f(\frac{2}{82})$, and so on. Stop. If you do that, you are walking into a trap.

In the world of JEE Advanced, we do not calculate; we observe. We look for the soul of the expression.

Let us look at the inputs. We are summing $f(\frac{k}{82})$ for $k=1,2,\dots ,81$. Look at the first term, where $k=1$, and the last term, where $k=81$.

The inputs are $\frac{1}{82}$ and $\frac{81}{82}$. What happens when we add them?

This is not a coincidence. This is a mathematical invitation. Whenever you see a series where the inputs sum to a constant, you must immediately test the symmetry property of the function. We are looking for the behavior of $f(x)+f(1-x)$.

Let us perform the algebra with precision. We have $f(x)=\frac{2^x}{2^x+\sqrt{2}}$. Now, let us construct $f(1-x)$ by replacing every $x$ with $1-x$:

This looks messy, but let us use the laws of indices. Recall that $2^{1-x}=\frac{2}{2^x}$. Substituting this into our expression, we get:

To clear this complex fraction, we multiply the numerator and the denominator by $2^x$. This yields:

Now, look at the denominator. We can factor out a $\sqrt{2}$. Since $2=(\sqrt{2}{)}^2$, we have:

Simplifying this, we get:

Now, the moment of truth. Let us add $f(x)$ and $f(1-x)$:

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

This is the 'Aha!' moment. The function is perfectly symmetric about $x=\frac{1}{2}$. Any two inputs that sum to $1$ will produce function values that sum to $1$.

We have $81$ terms in our summation. We can pair the first term ($k=1$) with the last term ($k=81$), the second term ($k=2$) with the second-to-last ($k=80$), and so on.

Since there are $81$ terms, we have $40$ pairs, and one term left in the middle. Each of these $40$ pairs sums to $1$. So, the sum of these $40$ pairs is $40\times 1=40$.

We cannot forget the middle term. The term that does not have a partner is the one where $k=41$. This gives us $f(\frac{41}{82})=f(\frac{1}{2})$.

Let us calculate this value:

Since $2^{1/2}=\sqrt{2}$, this becomes:

We have our $40$ from the pairs and our $\frac{1}{2}$ from the middle term. Adding them together:

Look at that. A massive, intimidating summation has collapsed into a simple fraction. This is the power of symmetry. When you approach a problem, do not just calculate; look for the structure, look for the symmetry, and let the math reveal its own elegance. You have mastered this problem. The final answer is $\frac{81}{2}$.