Animated Solution for Mathematics - Functions: A function $f(x)$ is given by $f(x)=\frac{5^x}{5^x+5}$, then the sum of the series $f(\frac{1}{20})+f(\frac{2}{20})+f(\frac{3}{20})+\cdots +f(\frac{39}{20})$ is equal to:

The problem presents the function $f(x)=\frac{5^x}{5^x+5}$ and a summation ranging from $x=\frac{1}{20}$ to $x=\frac{39}{20}$. Attempting to calculate each term individually is a strategic error.

In JEE Advanced, examiners prioritize identifying the hidden architecture of mathematics over brute-force arithmetic. We must look for patterns that simplify the expression.

Finding the Pattern

We must cultivate the Symmetry Mindset. Observe the inputs of the series: the first term is $\frac{1}{20}$ and the last is $\frac{39}{20}$.

Summing these extremes yields:

This is a clear signal. Whenever a series involves inputs that sum to a constant $a$, we must investigate the behavior of $f(x)+f(a-x)$. Here, our $a=2$.

We perform algebraic surgery by substituting $2-x$ into our function:

Applying exponent laws, we know $5^{2-x}=\frac{5^2}{5^x}=\frac{25}{5^x}$. Substituting this back into the function gives:

To simplify, multiply the numerator and the denominator by $5^x$:

Factoring out $5$ from the denominator:

Now, we calculate the sum $f(x)+f(2-x)$:

This is the "Aha!" moment. Every pair of terms whose inputs sum to $2$ will result in a sum of $1$.

The series contains $39$ terms. We can form $19$ pairs that each sum to $1$, totaling $19$. The middle term, $f(1)$, remains unpaired:

Adding the pairs to the middle term, we get $19+0.5=19.5$. The final result is $\frac{39}{2}$.

Remember: Symmetry is your greatest weapon in the JEE arsenal.