Let f x be a function such that f x + y = f x · f y for all x , y ∈ N , If f 1 = 3 and ∑ k = 1 n f k = 3279 , then the value of n is

Step-by-step solution

Given, f x + y = f x · f y and f 1 = 3 Now taking x = 1   &   y = 1 we get, f 1 + 1 = f 1 · f 1 ⇒ f 2 = 3 2 Similarly f 2 + 1 = f 2 · f 1 ⇒ f 3 = 3 2 × 3 = 3 3 And so on f n = 3 n So, ∑ r = 1 n f r = 3279 ⇒ f 1 + f 2 + f 3 . . . . . . . f n = 3279 ⇒ 3 + 3 2 + 3 3 . . . . . . . + 3 n = 3279 ⇒ 3 × 3 n - 1 3 - 1 = 3279 ⇒ 3 n - 1 2 = 1093 ⇒ 3 n - 1 = 2186 ⇒ 3 n = 2187 ⇒ 3 n = 3 7 ⇒ n = 7