Let f : R → R be a function which satisfies f x + y = f x + f y , ∀ x , y ∈ R . If f 1 = 2 and g n = ∑ k = 1 n - 1 f k , n ∈ N then the value of n , for which g n = 20 , is

Step-by-step solution

∵     f x + y = f x + f y ∴     f x = k x ,   ∀ x ∈ R Now f 1 = 2     ⇒   k = 2 ∴     f x = 2 x ,   ∀ x ∈ R g n = f 1 + f 2 + … … . . + f n - 1 = 2 1 + 2 + 3 + … … + n - 1 = 2 · n - 1 2 n = 20 ⇒ n 2 - n - 20 = 0 ⇒ n = 5 .