Let f x = 2 x 2 - x - 1 and S = n ∈ ℤ : f n ≤ 800 . Then, the value of ∑ n ∈ S f n is equal to _______.

Step-by-step solution

Given, f x = 2 x 2 - x - 1 and f n ≤ 800 So, 2 n 2 - n - 801 ≤ 0 ⇒ n 2 - 1 2 n - 801 2 ≤ 0 ⇒ n - 1 4 2 - 801 2 - 1 16 ≤ 0 ⇒ n - 1 4 2 - 6409 16 ≤ 0 ⇒ n - 1 4 - 6409 4 n - 1 4 + 6409 16 ≤ 0 ⇒ 1 - 6409 4 ≤ n ≤ 1 + 6409 4 Since given n is integer, So, n = - 19 , - 18 - 17 , … … . . 0 , 1 , 2 , … … , 20 Now ∑ n ∈ S f x = ∑ 2 x 2 - x - 1 = 2 ∑ x 2 - ∑ x - ∑ 1 = 2 19 2 + 18 2 + …