If the sum of the first 20 terms of the series 4.1/(4+3.1^2+1^4) + 4.2/(4+3.2^2+2^4) + 4.3/(4+3.3^2+

Q: Why did we add and subtract $r^2$ in the denominator?

This is a classic technique to factorize biquadratic expressions of the form $x^4 + ax^2 + b$. By adding and subtracting $r^2$, we create a perfect square $(r^2+2)^2$ and a difference of squares $(r^2+2)^2 - r^2$, which allows us to break the denominator into two manageable quadratic factors.

Q: How do I know if a series is telescoping?

Look for a general term that can be expressed as a difference of two similar functions, $f(r) - f(r+1)$. If you see that the terms in your sum start cancelling out (e.g., the second part of $T_1$ cancels with the first part of $T_2$), you are dealing with a telescoping series. It is one of the most elegant ways to sum a series!

Q: Why does the cancellation happen diagonally?

In a telescoping sum, the terms are structured such that the subtrahend of one term is the minuend of the next. When you write out the sum, the terms appear in a sequence where the negative part of $T_r$ is exactly the positive part of $T_{r+1}$. This creates a chain reaction of cancellation, leaving only the very first and very last components.

Q: Is there a shortcut for the last term?

While you can calculate the last term by plugging in $r=20$, the real 'shortcut' is understanding the structure of the telescoping sum. Once you have the form $2 \left( \frac{1}{f(r)} - \frac{1}{f(r+1)} \right)$, you know the sum will always be $2 \left( \frac{1}{f(1)} - \frac{1}{f(21)} \right)$. You don't need to write out every single term!

Question

If the sum of the first 20 terms of the series $\frac{4.1}{4+3.1^2+1^4} + \frac{4.2}{4+3.2^2+2^4} + \frac{4.3}{4+3.3^2+3^4} + \frac{4.4}{4+3.4^2+4^4} + \dots$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $m + n$ is equal to :-

Options:

(A) 423

(B) 420

(C) 421

(D) 422

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to dismantle a series that looks intimidating but is actually a masterpiece of algebraic symmetry.

When you first look at the series

$$ \sum_{r=1}^{20} \frac{4r}{r^4 + 3r^2 + 4} $$

it might seem like a chaotic mess. But in the world of JEE Advanced, chaos is just order waiting to be discovered. Let's embark on this journey to simplify it.

### The Anatomy of the Denominator

Look at the general term:

$$ T_r = \frac{4r}{r^4 + 3r^2 + 4} $$

The denominator, $D = r^4 + 3r^2 + 4$, is a biquadratic expression. If you try to solve it directly, you will hit a wall.

Instead, we use the method of completing the square. By adding and subtracting $r^2$, we transform it into $(r^4 + 4r^2 + 4) - r^2$, which is $(r^2+2)^2 - r^2$.

This is the key! It is a difference of squares: $(r^2-r+2)(r^2+r+2)$. This factorization is the foundation of our solution.

### The Art of Splitting

Now, look at the numerator. We have $4r$. Can we express $4r$ as the difference of the two factors we just found?

Let's check: $(r^2+r+2) - (r^2-r+2) = 2r$. Since our numerator is $4r$, we simply multiply this difference by $2$.

This is a strategic move, a classic JEE maneuver to force the expression into a telescoping form:

$$ T_r = 2 \left[ \frac{1}{r^2-r+2} - \frac{1}{r^2+r+2} \right] $$

### The Telescoping Magic

This is where the beauty of mathematics shines. As we sum these terms from $r=1$ to $20$, the middle terms vanish like magic.

For $r=1$, we have $2(\frac{1}{2} - \frac{1}{4})$. For $r=2$, we have $2(\frac{1}{4} - \frac{1}{8})$.

Notice how the $-\frac{1}{4}$ from the first term cancels with the $+\frac{1}{4}$ from the second? This dance of cancellation continues until the very end.

We are left with only the first positive component and the last negative component:

$$ S_{20} = 2 \left( \frac{1}{2} - \frac{1}{20^2 + 20 + 2} \right) = 2 \left( \frac{1}{2} - \frac{1}{422} \right) $$

### Final Calculation

Finally, we simplify the result:

$$ S_{20} = 2 \left( \frac{211 - 1}{422} \right) = 2 \left( \frac{210}{422} \right) = \frac{210}{211} $$

We are given that the sum is $\frac{m}{n}$ where $m$ and $n$ are coprime. Since $211$ is a prime number, $\gcd(210, 211) = 1$.

Thus, $m = 210$ and $n = 211$. The final step is to find $m+n$, which is $210 + 211 = 421$.

You have successfully navigated the complexity and found the elegant truth hidden within the series. The final answer is **421**.

Select Answer:

Visualized Solution

Identifying the General Term $T_{r}$

Analyzing the Denominator

Factorizing via Difference of Squares

Splitting the Numerator

Method of Differences (V-N Method)

Expanding the Series

The Last Term

Telescoping Cancellation

Simplifying the Result

Finding $m + n$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Anatomy of the Denominator

The Art of Splitting

The Telescoping Magic

Final Calculation

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