JEE Main 202621 Jan, Evening shiftMathematicsEasySequence and Series179× since 2002The positive integer n, for which the solutions of the equation x(x+2)+(x+2)(x+4)+⋯+(x+2n−2)(x+2n)=8n3x(x+2)+(x+2)(x+4)+\cdots+(x+2n-2)(x+2n)=\frac{8n}{3}x(x+2)+(x+2)(x+4)+⋯+(x+2n−2)(x+2n)=38n are two consecutive even integers, is :AOption A. 12BOption B. 3COption C. 6DOption D. 9CheckSkip · show solution
