The positive integer n, for which the solutions of the equation x(x+2) + (x+2)(x+4) + ... + (x+2n-2)

Q: Why can't we just expand the series manually for small values of n?

That is a classic trap! While you might find a pattern for $n=1$ or $n=2$, the JEE Advanced exam tests your ability to generalize. By using the summation notation $\sum_{r=1}^n T_r$, you are building a robust mathematical model that works for any $n$. It is the difference between guessing and proving.

Q: How do I know the roots are $2k$ and $2k+2$?

Great question! Even integers are always separated by a distance of 2. By defining them as $2k$ and $2k+2$, we capture the essence of 'consecutive even integers' perfectly. It simplifies the algebra significantly compared to using arbitrary variables.

Q: Why did we divide by $n$ in the middle of the calculation?

We divided by $n$ because the problem explicitly states that $n$ is a positive integer. This guarantees that $n \neq 0$. In algebra, dividing by a variable is dangerous unless you are absolutely certain it is non-zero. Here, we are safe, and it makes our quadratic equation much cleaner!

Question

The positive integer $n$, for which the solutions of the equation $x(x+2) + (x+2)(x+4) + \dots + (x+2n-2)(x+2n) = \frac{8n}{3}$ are two consecutive even integers, is :

Options:

(A) 3

(B) 12

(C) 9

(D) 6

Sigma Percentile · Accepted Answer

### The Symphony of Series and Roots: A JEE Masterclass

Welcome, fellow traveler on the path to JEE excellence. Today, we aren't just solving an equation; we are decoding a pattern.

When you first look at the expression

$$x(x+2) + (x+2)(x+4) + \dots + (x+2n-2)(x+2n) = \frac{8n}{3}$$

it is natural to feel a momentary surge of anxiety. It looks like a chaotic, never-ending string of terms, but in the world of JEE Advanced, chaos is often just order in disguise.

### Phase 1: Decoding the DNA of the Series

Imagine you are standing before a long, complex machine. You don't try to fix the whole machine at once; you look for the repeating component.

We define the general term, $T_r$. By observing the progression, we see that the first factor is $(x + 2r - 2)$ and the second is $(x + 2r)$.

So, our general term is:

$$T_r = (x + 2r - 2)(x + 2r)$$

When we expand this, we get:

$$T_r = x^2 + (4r - 2)x + 4r^2 - 4r$$

This is the 'DNA' of our series. Every single term in that long, intimidating sum is just a variation of this single expression. By focusing on $T_r$, we have already won half the battle.

### Phase 2: The Summation Battle

Now, we must aggregate these terms. We are looking for the sum $\sum_{r=1}^n T_r = \frac{8n}{3}$. We distribute the summation sign across our expanded $T_r$.

The first part is $\sum_{r=1}^n x^2$, which simply becomes $nx^2$ because $x^2$ is constant with respect to $r$.

The second part involves $\sum_{r=1}^n (4r - 2)x$. Using the standard sum of natural numbers formula, $\sum r = \frac{n(n+1)}{2}$, this simplifies to $2n^2x$.

The third part, $\sum_{r=1}^n (4r^2 - 4r)$, uses the sum of squares formula, $\sum r^2 = \frac{n(n+1)(2n+1)}{6}$.

After some algebraic heavy lifting, we arrive at a beautiful, simplified quadratic equation:

$$x^2 + 2nx + \frac{4n^2 - 12}{3} = 0$$

Take a moment to appreciate this. We have tamed the series and reduced it to a single, elegant quadratic form.

### Phase 3: The Geometric Intuition of Roots

Here is where the problem shifts from algebra to logic. We are told the roots, let's call them $\alpha$ and $\beta$, are consecutive even integers.

If they are consecutive even integers, their difference must be 2. Mathematically, $|\alpha - \beta| = 2$.

We know from Vieta's formulas that the sum of the roots is:

$$\alpha + \beta = -2n$$

And the product of the roots is:

$$\alpha\beta = \frac{4n^2 - 12}{3}$$

Let $\alpha = 2k$ and $\beta = 2k + 2$. Then $\alpha + \beta = 4k + 2 = -2n$, which implies $2k + 1 = -n$.

Now, look at the product: $\alpha\beta = (2k)(2k+2) = 4k^2 + 4k$. If we add 1 to this, we get $(2k+1)^2 = (-n)^2 = n^2$. Thus, $\alpha\beta = n^2 - 1$.

### Phase 4: The Final Victory

We now have two expressions for the product of the roots: $\alpha\beta = n^2 - 1$ and $\alpha\beta = \frac{4n^2 - 12}{3}$. Equating them is the final step in our journey:

$$n^2 - 1 = \frac{4n^2 - 12}{3}$$

Multiplying by 3 gives $3n^2 - 3 = 4n^2 - 12$. Rearranging this, we find $n^2 = 9$.

Since $n$ must be a positive integer, we discard the negative root and arrive at **$n = 3$**.

See how the complexity melted away? We didn't fight the problem; we understood its structure, applied the right tools, and let the math lead us to the truth. Keep this mindset, and no JEE problem will ever be too difficult for you.

Animated Solution for Mathematics - Sequence and Series: The positive integer $n$ , for which the solutions of the equation $x (x + 2) + (x + 2) (x + 4) + \dots + (x + 2 n - 2) (x + 2 n) = \frac{8 n}{3}$ are two consecutive even integers, is :

Select Answer:

Visualized Solution

Analyze the Series Pattern

Define the General Term $T_{r}$

Expand the General Term

Set up the Summation

Summing the Terms

The Final Quadratic Form

Analyze the Roots

Sum of Roots Property

Product of Roots Property

Equate and Solve for $n$

Final Answer

The Sigma Insight: Sum of Special Series

The Symphony of Series and Roots

Phase 1

Phase 2

Phase 3

Phase 4

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