Animated Solution for Mathematics - Sequence and Series: The sum of $10$ terms of the series $\frac{3}{1^2\times 2^2}+\frac{5}{2^2\times 3^2}+\frac{7}{3^2\times 4^2}+\dots$ is :

A JEE Masterclass

Welcome, future engineer. Today, we are not just solving a math problem; we are embarking on a journey to uncover the hidden elegance of series summation.

When you first look at the series $\frac{3}{1^2\times 2^2}+\frac{5}{2^2\times 3^2}+\frac{7}{3^2\times 4^2}+\dots$, it is natural to feel a bit overwhelmed. You might be tempted to start calculating $1^2\times 2^2$, then $2^2\times 3^2$, and so on.

But stop! In the JEE Advanced arena, if you find yourself doing heavy arithmetic, you are likely missing the 'elegant path'. Let us find that path together.

The Detective Work

Every series has a heartbeat, and that heartbeat is the general term, $T_r$. To find it, we must be detectives.

Look at the numerators: $3,5,7,\dots$. This is an arithmetic progression with a first term of $3$ and a common difference of $2$. Thus, the $r$-th numerator is $3+(r-1)2=2r+1$.

Now, look at the denominators: $(1^2\cdot 2^2),(2^2\cdot 3^2),(3^2\cdot 4^2),\dots$. The pattern is undeniable. The $r$-th term's denominator is $r^2(r+1{)}^2$.

So, our general term is:

This is the raw material we need to transform.

The Algebraic Alchemy

Now, here is the secret that separates the masters from the novices. We need to manipulate $T_r$ into a form that allows for cancellation.

Look at the denominator factors: $r^2$ and $(r+1{)}^2$. What happens if we subtract them? Let's expand $(r+1{)}^2-r^2$.

We get $(r^2+2r+1)-r^2$, which simplifies beautifully to $2r+1$. Do you see it? The numerator is exactly the difference of the two squares in the denominator!

This is not a coincidence; it is the design of the problem. We can rewrite our general term as:

By splitting this fraction, we get:

Canceling the common terms, we arrive at the simplified, elegant form:

The Telescoping Collapse

We are now ready for the grand finale. We need the sum of the first $10$ terms, $S_{10}={\sum }_{r=1}^{10}{T}_r$.

Substituting our new form, we have:

Let's expand this sum to witness the magic. For $r=1$, we have $(\frac{1}{1^2}-\frac{1}{2^2})$. For $r=2$, we have $(\frac{1}{2^2}-\frac{1}{3^2})$. For $r=3$, we have $(\frac{1}{3^2}-\frac{1}{4^2})$.

This continues until $r=10$, where we have $(\frac{1}{10^2}-\frac{1}{11^2})$. When we add these together, the $-\frac{1}{2^2}$ from the first term cancels with the $+\frac{1}{2^2}$ from the second.

The $-\frac{1}{3^2}$ cancels with the $+\frac{1}{3^2}$, and so on. Like a telescope collapsing, all intermediate terms vanish, leaving only the first part of the first term and the last part of the last term:

We are left with $1-\frac{1}{121}$. Calculating this is straightforward:

There you have it! We didn't need to calculate ten complex fractions. We used the power of pattern recognition and algebraic manipulation to collapse the problem into a simple subtraction.

This is the mindset of a JEE topper: always look for the structure, always look for the simplification, and never be afraid to manipulate the expression to reveal its true nature. Keep practicing, and soon, you will see these patterns instantly!