Animated Solution for Mathematics - Sequence and Series: $\frac{1}{3^2-1}+\frac{1}{5^2-1}+\frac{1}{7^2-1}+\cdots +\frac{1}{(201{)}^2-1}$ is equal to

Mastering the Telescoping Series

Welcome, future engineer. Today, we are not just solving a math problem; we are embarking on a journey of pattern recognition.

When you first look at the expression $\frac{1}{3^2-1}+\frac{1}{5^2-1}+\frac{1}{7^2-1}+\cdots +\frac{1}{201^2-1}$, it is natural to feel a slight tremor of intimidation. It looks like a mountain of arithmetic.

But in the world of JEE Advanced, we do not climb mountains by brute force; we find the path of least resistance. We find the elegance hidden within the chaos.

The Hidden Pattern

Let us pause and observe. The denominators are $3^2-1$, $5^2-1$, $7^2-1$, and so on. If you calculate these, you get $8,24,48$, and so on.

That does not look very helpful, does it? But look closer at the bases: $3,5,7,\dots ,201$. These are consecutive odd numbers.

This is the heartbeat of our problem. If we can define the $n^{th}$ term, we can conquer the entire series.

We know that any odd number can be represented as $2n+1$ (starting from $n=1$, which gives $3$). So, our general term $T_n$ is defined as:

The Algebraic Key

Now, here is where most students stumble. They try to expand the square: $(2n+1{)}^2=4n^2+4n+1$. Then they subtract $1$, getting $4n^2+4n$.

While this is correct, it is a dead end. It obscures the structure. Instead, remember the most powerful tool in your algebraic arsenal: the difference of squares identity, $a^2-b^2=(a-b)(a+b)$.

Let $a=2n+1$ and $b=1$. Our denominator becomes:

Look at the beauty of this simplification! The first bracket becomes $2n$, and the second becomes $2n+2$. Our general term is now:

If we factor out a $2$ from the second term, we get $2(n+1)$. Thus, our term simplifies to:

The Method of Differences

We have arrived at a pivotal moment. We have $\frac{1}{n(n+1)}$. This is the signature of a telescoping series.

We want to split this into a difference of two fractions. Notice that the difference between the factors in the denominator is $(n+1)-n=1$. This is exactly our numerator!

We can rewrite the numerator $1$ as $(n+1)-n$. Therefore:

Applying the constant $\frac{1}{4}$ we found earlier, our general term is now:

The Final Collapse

Before we sum this up, we must know how many terms we are dealing with. The last term has a base of $201$.

Setting $2n+1=201$, we find $2n=200$, so $n=100$. We are summing from $n=1$ to $n=100$.

Now, watch the magic. Let us write out the sum $S$:

Expanding this, we get:

Do you see it? The $-\frac{1}{2}$ cancels with the $+\frac{1}{2}$. The $-\frac{1}{3}$ cancels with the $+\frac{1}{3}$.

Like a telescope collapsing into itself, every intermediate term vanishes. Only the very first term ($1$) and the very last term ($-\frac{1}{101}$) survive the carnage.

We are left with a simple calculation:

And there it is. A problem that looked like a nightmare of arithmetic has been reduced to an elegant, simple fraction.

This is the power of mathematical thinking. You didn't just calculate; you understood the structure. Keep this mindset, and no problem will ever be too big for you.