JEE Main 2026 (Online) 21st January Evening ShiftQuestion 1
The positive integer $n$, for which the solutions of the equation $x(x+2)+(x+2)(x+4)+\cdots+(x+2n-2)(x+2n)=\frac{8n}{3}$ are two consecutive even integers, is :
Sequence and Series · Standard Sums (Σ formulas)
Solution
Formulas used. (1) $\sum_{k=0}^{n-1}k=\frac{n(n-1)}{2}$ and $\sum_{k=0}^{n-1}k(k+1)=\frac{(n-1)n(n+1)}{3}$(2) Roots of $x^2+px+q=0$ differ by $\sqrt{p^2-4q}$Step 1 - General term. The $k$-th product ($k=0,1,\dots,n-1$) is $$(x+2k)(x+2k+2)=x^2+(4k+2)x+4k(k+1).$$ Step 2 - Add the $n$ terms. $$nx^2+x\left[4\cdot\frac{n(n-1)}{2}+2n\right]+4\cdot\frac{(n-1)n(n+1)}{3}=nx^2+2n^2x+\frac{4n(n^2-1)}{3}.$$ Step 3 - The equation. Set this equal to $\dfrac{8n}{3}$ and divide by $n$: $$x^2+2nx+\frac{4n^2-4-8}{3}=0\ \Longrightarrow\ x^2+2nx+\frac{4n^2-12}{3}=0 .$$ Step 4 - Roots are consecutive even integers, so they differ by $2$: $$(2n)^2-4\cdot\frac{4n^2-12}{3}=2^2\ \Longrightarrow\ 3n^2-(4n^2-12)=3\ \Longrightarrow\ n^2=9,\ n=3 .$$ Check. For $n=3$: $x^2+6x+8=0$, roots $-2,-4$, two consecutive even integers. Answer: (A)