If Σ (k=1 to 10) k / (k⁴ + k² + 1) = m/n, where m and n are co-prime, then m + n is equal to _______

Q: Wait, how was I supposed to know to add and subtract $k^2$ in the denominator? That feels like a random trick!

It is not random, it is a classic pattern! In JEE Advanced, whenever you see a quartic polynomial of the form $x^4 + ax^2 + 1$, your first instinct should be to complete the square. It is a standard technique to transform a difficult expression into a product of two quadratics. Practice makes this 'trick' feel like second nature.

Q: Why does the telescoping series work so perfectly? Is it always guaranteed to cancel out like this?

Telescoping series rely on the structure $f(k) - f(k+1)$. If you can manipulate your general term into this specific difference, the intermediate terms will always cancel. It is not magic; it is the beauty of algebraic symmetry. If you find yourself with a sum where terms don't cancel, re-check your partial fraction decomposition!

Q: Why do we reject the negative root or worry about co-prime conditions?

The problem specifically asks for $m$ and $n$ to be co-prime to ensure a unique representation of the fraction. If you didn't simplify your fraction to its lowest terms, you would get a different $m+n$. Always check the GCD of your final numerator and denominator!

Question

If $\sum_{k=1}^{10} \frac{k}{k^4 + k^2 + 1} = \frac{m}{n}$, where $m$ and $n$ are co-prime, then $m + n$ is equal to ________.

Sigma Percentile · Accepted Answer

### The Beauty of the Hidden Structure

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of powers and summations.

We are looking at the series:

$$\sum_{k=1}^{10} \frac{k}{k^4 + k^2 + 1}$$

When you see a denominator like $k^4 + k^2 + 1$, your instinct might be to panic. It is a quartic polynomial, and we rarely want to deal with those directly. But in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask.

### Phase 1: The Denominator Mystery

Our first task is to tame the denominator. We have $k^4 + k^2 + 1$. This is a classic form.

If we treat this as a quadratic in terms of $k^2$, we can complete the square. Watch closely: if we add $k^2$ and subtract $k^2$, we get:

$$(k^4 + 2k^2 + 1) - k^2$$

Suddenly, the first three terms collapse into a perfect square: $(k^2 + 1)^2 - k^2$. Now, we have a difference of two squares, $a^2 - b^2$, which we know factors into $(a-b)(a+b)$.

Applying this, our denominator becomes:

$$(k^2 - k + 1)(k^2 + k + 1)$$

This is the key that unlocks the entire problem.

### Phase 2: The Art of Decomposition

Now that we have factored the denominator, we look at our general term:

$$T_k = \frac{k}{(k^2 - k + 1)(k^2 + k + 1)}$$

We need to split this into partial fractions. Instead of using the standard method of undetermined coefficients, let us use observation.

Look at the two factors in the denominator: $(k^2 + k + 1)$ and $(k^2 - k + 1)$. What happens if we subtract the smaller from the larger?

$$(k^2 + k + 1) - (k^2 - k + 1) = 2k$$

This is brilliant! Our numerator is $k$, which is exactly half of this difference. So, we can rewrite $T_k$ as:

$$T_k = \frac{1}{2} \left[ \frac{(k^2 + k + 1) - (k^2 - k + 1)}{(k^2 - k + 1)(k^2 + k + 1)} \right]$$

This simplifies beautifully to:

$$T_k = \frac{1}{2} \left[ \frac{1}{k^2 - k + 1} - \frac{1}{k^2 + k + 1} \right]$$

### Phase 3: The Telescoping Collapse

Now, we unleash the power of the telescoping series. We are summing from $k=1$ to $k=10$.

Let us write out the first few terms:
For $k=1$, we have $\frac{1}{2} [ \frac{1}{1} - \frac{1}{3} ]$. 
For $k=2$, we have $\frac{1}{2} [ \frac{1}{3} - \frac{1}{7} ]$. 
For $k=3$, we have $\frac{1}{2} [ \frac{1}{7} - \frac{1}{13} ]$.

Do you see the chain reaction? The negative term of one bracket cancels the positive term of the next.

This is the 'telescoping' effect—like an old-fashioned collapsible telescope, the entire series collapses into just the first part of the first term and the last part of the last term. The intermediate terms are all destroyed in this beautiful, mathematical crossfire.

### The Final Victory

After the dust settles, we are left with:

$$\frac{1}{2} \left[ 1 - \frac{1}{10^2 + 10 + 1} \right] = \frac{1}{2} \left[ 1 - \frac{1}{111} \right]$$

This simplifies to:

$$\frac{1}{2} \left[ \frac{110}{111} \right] = \frac{55}{111}$$

We are told this is $\frac{m}{n}$ where $m$ and $n$ are co-prime. Since $55$ and $111$ share no common factors, we have $m=55$ and $n=111$.

The final step is simply $m + n = 55 + 111 = 166$. You have navigated the complexity, utilized the algebraic identity, and mastered the telescoping series. This is the essence of JEE Advanced—not just solving, but seeing the structure beneath the surface.

Animated Solution for Mathematics - Sequence and Series: If $\sum_{k = 1}^{10} \frac{k}{k ^{4} + k ^{2} + 1} = \frac{m}{n}$ , where $m$ and $n$ are co-prime, then $m + n$ is equal to ________.

Enter Numerical Value:

Visualized Solution

Identifying the General Term $T_{k}$

The Denominator Trick

Factorizing the Denominator

Method of Differences Setup

Partial Fraction Decomposition

The Telescoping Series

Cancellation Pattern

Simplifying the Result

Finding $m + n$

The Sigma Insight: Sum of Special Series

The Beauty of the Hidden Structure

Phase 1

Phase 2

Phase 3

The Final Victory

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑k=110​k4+k2+1k​=nm​, where m and n are co-prime, then m+n is equal to ________.