Let , + 2, 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) = 4n for some n N . Then n + is equal to :

Step-by-step solution

The given equation is: _ k=0 ^ n-1 (x+k)(x+k+2) = 4n Expanding the terms inside the summation: _ k=0 ^ n-1 (x^2 + 2(k+1)x + k(k+2)) = 4n Summing each term separately: n x^2 + 2x _ k=0 ^ n-1 (k+1) + _ k=0 ^ n-1 (k^2 + 2k) = 4n Using the standard summation formulas: n x^2 + 2x n(n+1) 2 + (n-1)n(2n-1) 6 + 2 (n-1)n 2 = 4n n x^2 + n(n+1)x + n(n-1)(2n+5) 6 = 4n Since n N , dividing the entire equation by n gives: x^2 + (n+1)x + (n-1)(2n+5) 6 - 4 = 0 The roots of this quadratic equation are given as and + 2 . Using the su