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JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 1

Let α, α+2\alpha,\ \alpha+2, α∈Z\alpha\in\mathbb Z, be the roots of the quadratic equation
x(x+2)+(x+1)(x+3)+(x+2)(x+4)+⋯+(x+n−1)(x+n+1)=4nx(x+2)+(x+1)(x+3)+(x+2)(x+4)+\dots+(x+n-1)(x+n+1)=4n
for some n∈Nn\in\mathbb N. Then n+αn+\alpha is equal to :

Choose one option
Answer and worked solution

Official answer: C

Expand. The sum is ∑k=0n−1(x+k)(x+k+2)=nx2+(n2+n)x+∑k=0n−1k(k+2)\sum_{k=0}^{n-1}(x+k)(x+k+2)=nx^2+(n^2+n)x+\sum_{k=0}^{n-1}k(k+2), and ∑k=0n−1k(k+2)=n(n−1)(2n+5)6\sum_{k=0}^{n-1}k(k+2)=\frac{n(n-1)(2n+5)}6. Dividing by nn:
x2+(n+1)x+(n−1)(2n+5)6−4=0.x^2+(n+1)x+\frac{(n-1)(2n+5)}6-4=0.

Roots differ by 2, so the discriminant is 4:
(n+1)2−4[(n−1)(2n+5)6−4]=4  ⟹  49−n2=0  ⟹  n=7.(n+1)^2-4\left[\frac{(n-1)(2n+5)}6-4\right]=4\implies49-n^2=0\implies n=7.

Roots. Sum 2α+2=−(n+1)=−82\alpha+2=-(n+1)=-8, so α=−5\alpha=-5 (roots −5, −3). Then n+α=2n+\alpha=2.

(C)

Topics:Roots and coefficientsSpecial series sums
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