Suppose f : R → 0 , ∞ be a differentiable function such that 5 f x + y = f x · f y , ∀ x , y ∈ R , If f 3 = 320 , then ∑ n = 0 5 f n is equal to:

Step-by-step solution

Given, equation is: 5 f x + y = f x · f y       . . . 1 At x = 0 ,   y = 0 , we get 5 f 0 = f 0 2 ⇒ f 0 = 5 Put y = 1 in 1 , then we get 5 f x + 1 = f x · f 1 ⇒ f x + 1 f x = f 1 5 ⇒ f 1 f 0 · f 2 f 1 · f 3 f 2 = f 1 5 3 ⇒ 320 5 = f 1 3 5 3 ⇒ f 1 = 20 ∴   5 f x + 1 = 20 · f x ⇒ f x + 1 = 4 f x So, f 1 = 4 f 0 = 4 · 5 f 2 = 4 f 1 = 4 2 · 5 f 3 = 4 f 2 = 4 3 · 5 f 4 = 4 f 3 = 4 4 · 5 f 5 = 4 f 4 = 4 5 &#183