JEE Main 2026 — Mathematics Algebra

The given equation is:

$$\sum_{k=0}^{n-1} (x+k)(x+k+2) = 4n$$k=0∑n−1​(x+k)(x+k+2)=4n

Expanding the terms inside the summation:

$$\sum_{k=0}^{n-1} (x^2 + 2(k+1)x + k(k+2)) = 4n$$k=0∑n−1​(x2+2(k+1)x+k(k+2))=4n

Summing each term separately:

$$n x^2 + 2x \sum_{k=0}^{n-1} (k+1) + \sum_{k=0}^{n-1} (k^2 + 2k) = 4n$$nx2+2xk=0∑n−1​(k+1)+k=0∑n−1​(k2+2k)=4n

Using the standard summation formulas:

$$n x^2 + 2x \dfrac{n(n+1)}{2} + \dfrac{(n-1)n(2n-1)}{6} + 2 \dfrac{(n-1)n}{2} = 4n$$nx2+2x2n(n+1)​+6(n−1)n(2n−1)​+22(n−1)n​=4n

$$n x^2 + n(n+1)x + \dfrac{n(n-1)(2n+5)}{6} = 4n$$nx2+n(n+1)x+6n(n−1)(2n+5)​=4n

Since $n \in \mathbb{N}$n∈N, dividing the entire equation by $n$n gives:

$$x^2 + (n+1)x + \dfrac{(n-1)(2n+5)}{6} - 4 = 0$$x2+(n+1)x+6(n−1)(2n+5)​−4=0

The roots of this quadratic equation are given as $\alpha$α and $\alpha + 2$α+2.

Using the sum of the roots:

$$\alpha + (\alpha + 2) = -(n+1)$$α+(α+2)=−(n+1)

$$2\alpha + 2 = -n - 1 \Rightarrow n = -2\alpha - 3$$2α+2=−n−1⇒n=−2α−3

Using the product of the roots:

$$\alpha(\alpha + 2) = \dfrac{(n-1)(2n+5)}{6} - 4$$α(α+2)=6(n−1)(2n+5)​−4

Substituting $n = -2\alpha - 3$n=−2α−3 into the product equation:

$$\alpha^2 + 2\alpha = \dfrac{(-2\alpha - 4)(-4\alpha - 1)}{6} - 4$$α2+2α=6(−2α−4)(−4α−1)​−4

$$6(\alpha^2 + 2\alpha) = 8\alpha^2 + 18\alpha + 4 - 24$$6(α2+2α)=8α2+18α+4−24

$$6\alpha^2 + 12\alpha = 8\alpha^2 + 18\alpha - 20$$6α2+12α=8α2+18α−20

$$2\alpha^2 + 6\alpha - 20 = 0$$2α2+6α−20=0

$$\alpha^2 + 3\alpha - 10 = 0$$α2+3α−10=0

$$(\alpha + 5)(\alpha - 2) = 0$$(α+5)(α−2)=0

Thus, $\alpha = -5$α=−5 or $\alpha = 2$α=2.

If $\alpha = 2$α=2, then $n = -2(2) - 3 = -7$n=−2(2)−3=−7, which is rejected since $n \in \mathbb{N}$n∈N.

If $\alpha = -5$α=−5, then $n = -2(-5) - 3 = 7$n=−2(−5)−3=7, which is valid.

Therefore, $n + \alpha = 7 + (-5) = 2$n+α=7+(−5)=2.

Answer: $2$2