Let ∑ k = 1 10 f a + k = 16 2 10 - 1 , where the function f satisfies f x + y = f x f y for all natural numbers x , y and f 1 = 2 . Then the natural number ' a ' is:

Step-by-step solution

Given f 1 = 2 and f x + y = f x · f y Put x = y = 1 ⇒ f 2 = f 1 · f 1 = 2 2 Now, put x = 1 ,   y = 2 ⇒ f 3 = f 1 · f 2 = 2 3 Now, put x = 1 ,   y = 3 ⇒ f 4 = f 1 · f 3 = 2 4 Similarly, f 10 = 2 10 Now ∑ k = 1 10   f a + k = 16 2 10 - 1 ⇒ ∑ k = 1 10 f a f k = 16 2 10 - 1 ⇒ f a ∑ k = 1 10 f k = 16 2 10 - 1 ⇒ f a f 1 + f 2 … + f 10 = 16 2 10 - 1 ⇒ f a 2 1 + 2 2 + 2 3 + … + 2 10 = 16 2 10 - 1 Using the sum of n te