Animated Solution for Mathematics - Sequence and Series: The sum ${\sum }_{n=1}^{21}\frac{3}{(4n-1)(4n+3)}$ is equal to

Welcome, future engineer. Today, we are going to peel back the layers of a problem that often intimidates students at first glance: the summation of a series.

When you see an expression like

your instinct might be to panic. But take a breath; this is not a monster, but a puzzle waiting to be solved.

The core of this problem lies in the denominator, which is a product of two linear factors: $(4n-1)$ and $(4n+3)$. In the world of JEE Advanced, whenever you see a product of linear factors in the denominator, your mind should immediately jump to the Method of Differences. This is our secret weapon.

Let us analyze the difference between these two factors. If we subtract $(4n-1)$ from $(4n+3)$, we get:

This constant difference of $4$ is the golden ticket. It allows us to rewrite the numerator.

We currently have a $3$, but we need a $4$. So, we multiply and divide by $4$, giving us:

Now, we replace that $4$ with the difference $(4n+3)-(4n-1)$. This transforms our term into:

This is the magic of the telescoping series.

As we expand the sum from $n=1$ to $n=21$, we see a beautiful pattern emerge.

For $n=1$, we have $\frac{3}{4}\left(\frac{1}{3}-\frac{1}{7}\right)$. For $n=2$, we have $\frac{3}{4}\left(\frac{1}{7}-\frac{1}{11}\right)$.

Notice how the $-\frac{1}{7}$ from the first term cancels with the $+\frac{1}{7}$ from the second term? This continues all the way to the end.

The intermediate terms vanish, leaving only the first part of the first term and the last part of the last term. We are left with:

Calculating this, we get:

Simplifying the expression, we arrive at the final result:

This is the elegance of mathematics. What seemed complex is actually a simple, beautiful collapse. Keep practicing, keep questioning, and you will master these concepts.