The positive integer n, for which the solutions of the equation x(x+2)+(x+2)(x+4)+ +(x+2 n-2)(x+2 n)= 8 n 3 are two consecutive even integers, is :

Step-by-step solution

The sum has n terms: _ k=1 ^ n (x+2k-2)(x+2k) . Expanding and summing gives nx^2 + 2n^2x + 4n(n^2-1) 3 = 8n 3 . Dividing by n and simplifying: 3x^2 + 6nx + 4n^2 - 12 = 0 . Discriminant = 36n^2 - 12(4n^2 - 12) = 12(12 - n^2) . Difference of roots = 2 3(12-n^2) 3 . For two consecutive even integers, this difference = 2 . 36 - 3n^2 = 3 n^2 = 9 n = 3 . For n = 3 : x = -9 3 3 = -2 or -4 , which are consecutive even integers.