Let a , b , c ∈ R . If f x = a x 2 + b x + c is such that a + b + c = 3 and f x + y = f x + f y + x y , ∀ x , y ∈ R , then ∑ n = 1 10 f ( n ) is equal to:

Step-by-step solution

Given, f x + y = f x + f y + x y     ∀ x ,   y ∈ R Put y = 1 , we get ⇒ f x + 1 = f x + f 1 + x ⇒ f x + 1 - f x = 3 + x On taking Sigma on both sides ⇒   ∑ x = 1 n - 1 f x + 1 - f x = ∑ x = 1 n - 1 3 + x ⇒ f 2 - f 1 + f 3 - f 2 + f 4 - f 3 + f 5 - f 4 + . . . . . . + f n - f n - 1 = 3 n - 1 + n - 1 n 2 ⇒ f n - f 1 = 3 n - 1 + n - 1 n 2 ⇒ f n - 3 = 3 n - 1 + n - 1 n 2 ⇒ f n = n 2 2 + 5 n 2 Again taking Sigma on both the sides &#865