If Σ (r=1 to 25) r/(r⁴+r²+1) = p/q, where p and q are positive integers such that gcd(p, q) = 1, the

Q: Wait, why do we reject the negative root here? Or why can't we just use energy conservation?

This is a pure algebraic series problem, not a physics problem! There are no roots to reject because we are not solving a quadratic equation for a variable. We are simply manipulating the expression to reveal a telescoping structure. Stay focused on the algebraic pattern!

Q: How do I know when to use the telescoping series method?

Great question! Whenever you see a rational function in a summation where the denominator is a polynomial of degree 4 or higher, and the numerator is of a lower degree, always check if the denominator can be factored into two parts whose difference is proportional to the numerator. That is the hallmark of a telescoping series.

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

We did that to complete the square! By writing $r^4+r^2+1$ as $(r^4+2r^2+1)-r^2$, we transformed it into $(r^2+1)^2 - r^2$, which is a difference of squares. This is a standard technique to factorize quartic polynomials that don't have obvious roots.

Question

If $\sum_{r=1}^{25} \frac{r}{r^4+r^2+1} = \frac{p}{q}$, where $p$ and $q$ are positive integers such that $\gcd(p, q) = 1$, then $p + q$ is equal to ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the fascinating world of series and sequences. Today, we are going to tackle a problem that might look intimidating at first glance, but beneath its complex exterior lies a structure of pure, elegant simplicity.

We are looking at the summation:

$$\sum_{r=1}^{25} \frac{r}{r^4+r^2+1} = \frac{p}{q}$$

At first, you might be tempted to start plugging in values for $r=1, 2, 3...$, but stop! That is a trap. In JEE Advanced, we do not brute force; we look for the hidden architecture.

The key here is the denominator: $r^4+r^2+1$. This is a classic expression that screams for a specific algebraic manipulation. If we add and subtract $r^2$, we get $(r^4+2r^2+1) - r^2$, which is $(r^2+1)^2 - r^2$.

Now, using the difference of squares identity $a^2-b^2 = (a-b)(a+b)$, we can factor this into $(r^2-r+1)(r^2+r+1)$. This is our breakthrough!

### The Numerator Trick

Now that we have our factors, let us look at the numerator, $r$. We want to express $r$ in terms of our new factors, $(r^2-r+1)$ and $(r^2+r+1)$.

If we subtract the smaller factor from the larger one, we get $(r^2+r+1) - (r^2-r+1) = 2r$. This is perfect! It means we can write:

$$r = \frac{1}{2} [(r^2+r+1) - (r^2-r+1)]$$

By substituting this back into our general term $T_r$, we get:

$$T_r = \frac{\frac{1}{2} [(r^2+r+1) - (r^2-r+1)]}{(r^2+r+1)(r^2-r+1)}$$

When we split this into two fractions, the terms cancel out beautifully, leaving us with:

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

### The Telescoping Collapse

This is where the magic happens. Let us define a function $f(r) = \frac{1}{r^2-r+1}$. If you calculate $f(r+1)$, you will find it is exactly:

$$f(r+1) = \frac{1}{(r+1)^2-(r+1)+1} = \frac{1}{r^2+r+1}$$

Our general term is now $T_r = \frac{1}{2} [f(r) - f(r+1)]$. When we sum this from $r=1$ to $25$, we get:

$$\frac{1}{2} [(f(1)-f(2)) + (f(2)-f(3)) + \dots + (f(25)-f(26))]$$

Notice the pattern? The $-f(2)$ cancels with $+f(2)$, the $-f(3)$ cancels with $+f(3)$, and so on. This is the 'telescoping' effect—the series collapses like a folding telescope, leaving only the first and last terms:

$$\frac{1}{2} [f(1) - f(26)]$$

### The Final Victory

Now, we just need to calculate the values. $f(1) = \frac{1}{1^2-1+1} = 1$.

For $f(26)$, we have:

$$f(26) = \frac{1}{26^2-26+1} = \frac{1}{676-26+1} = \frac{1}{651}$$

Our sum is:

$$\frac{1}{2} \left[ 1 - \frac{1}{651} ight] = \frac{1}{2} \left[ \frac{650}{651} ight] = \frac{325}{651}$$

We are given that the sum is $\frac{p}{q}$ with $\gcd(p, q) = 1$. Since $325 = 5^2 	imes 13$ and $651 = 3 	imes 7 	imes 31$, they are indeed coprime.

Thus, $p=325$ and $q=651$. The final answer, **$p+q$**, is **$325+651 = 976$**.

Animated Solution for Mathematics - Sequence and Series: If $\sum_{r = 1}^{25} \frac{r}{r ^{4} + r ^{2} + 1} = \frac{p}{q}$ , where $p$ and $q$ are positive integers such that $g cd (p, q) = 1$ , then $p + q$ is equal to ......... .

Enter Numerical Value:

Visualized Solution

Analyze the General Term $T_{r}$

Factorizing the Denominator

Decomposing the Numerator

Partial Fraction Form

Identifying the Telescoping Pattern

Expanding the Summation

Calculating $f (1)$ and $f (26)$

Final Sum Calculation

Finding $p + q$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Numerator Trick

The Telescoping Collapse

The Final Victory

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