JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Quadratic Equations and Inequalities

Each term has the form $(x+k)(x+k+2)$ for $k=0,1,\dots ,n-1$. Summing, ${\sum }_{k=0}^{n-1}(x+k)(x+k+2)=n{x}^2+n(n+1)x+\frac{n(n-1)(2n+5)}{6}$, so the equation $=4n$ reduces, on dividing by $n$, to $x^2+(n+1)x+\frac{(n-1)(2n+5)}{6}-4=0$. Its roots are $\alpha$ and $\alpha +2$, which differ by $2$; hence $(\text{difference}{)}^2=(\text{sum}{)}^2-4(\text{product})=4$, i.e. $(n+1{)}^2-4\left[\frac{(n-1)(2n+5)}{6}-4\right]=4$. This simplifies to $-n^2+61=12$, so $n^2=49$ and $n=7$. For $n=7$ the equation becomes $x^2+8x+15=0$ with roots $-5$ and $-3$, giving $\alpha =-5$. Therefore $n+\alpha =7+(-5)=2$.