Animated Solution for Mathematics - Sequence and Series: If $a_n=\frac{-2}{4n^2-16n+15}$, then $a_1+a_2+\cdots +a_{25}$ is equal to :

Unlocking the Hidden Pattern

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dismantle a problem that, at first glance, looks like a tedious, soul-crushing arithmetic slog.

You see a sum of 25 terms, and your instinct might be to panic. But wait—in the world of competitive mathematics, whenever you see a complex rational expression inside a summation, there is almost always a hidden, elegant structure waiting to be revealed. Let us peel back the layers together.

The Anatomy of the Denominator

We are given the general term $a_n=\frac{-2}{4n^2-16n+15}$. Our mission is to compute the sum $S_{25}={\sum }_{n=1}^{25}{a}_n$.

If we were to calculate each term individually, we would be here until the next century. Instead, let us look at the denominator: $4n^2-16n+15$. This is a quadratic expression.

Let us factorize it by splitting the middle term. We look for two numbers that multiply to $4\times 15=60$ and add to $-16$. Those numbers are $-10$ and $-6$.

Thus, we rewrite the expression as $4n^2-10n-6n+15$. Grouping them, we get $2n(2n-5)-3(2n-5)$, which simplifies beautifully to $(2n-3)(2n-5)$.

The Magic of Partial Fractions

Now, look at the numerator. It is a constant $-2$. Is it a coincidence that the difference between our factors $(2n-3)$ and $(2n-5)$ is exactly $2$? No, it is not!

This is the 'Aha!' moment. We can rewrite the numerator $-2$ as $(2n-5)-(2n-3)$. By substituting this back into our expression for $a_n$, we get:

When we split this fraction, we arrive at the most powerful form of the term: $a_n=\frac{1}{2n-3}-\frac{1}{2n-5}$. This, my friend, is the key to the kingdom. We have transformed a complex fraction into a difference of two simpler terms.

The Telescoping Collapse

Imagine a row of dominoes. When we write out the terms of our sum, something magical happens. Let us look at the first few terms:

For $n=1$: $a_1=\frac{1}{-1}-\frac{1}{-3}$

For $n=2$: $a_2=\frac{1}{1}-\frac{1}{-1}$

For $n=3$: $a_3=\frac{1}{3}-\frac{1}{1}$

Do you see it? The term $-\frac{1}{-1}$ in $a_1$ is cancelled by the $+\frac{1}{-1}$ in $a_2$. The term $-\frac{1}{1}$ in $a_2$ is cancelled by the $+\frac{1}{1}$ in $a_3$.

This is the 'telescoping' effect—like an old-fashioned pirate's telescope collapsing into itself. Every term in the middle is destined to vanish, leaving only the 'start' of the first term and the 'end' of the last term.

The Final Victory

We only need to evaluate the very last term, $a_{25}$. Plugging $n=25$ into our decomposed form, we get $a_{25}=\frac{1}{2(25)-3}-\frac{1}{2(25)-5}$, which is $\frac{1}{47}-\frac{1}{45}$.

Following our pattern, the surviving terms are the second part of $a_1$ and the first part of $a_{25}$. Our sum is simply:

To finish, we find a common denominator:

Look at that! We started with a daunting quadratic fraction and ended with a simple, elegant fraction. This is the beauty of mathematics—it rewards those who look for patterns rather than those who blindly calculate. You have mastered the telescoping series today. Keep this intuition sharp; it will serve you well in the toughest battles of the JEE Advanced.