Question

Find the natural number a for which ` sum_(k = 1)^n f(a + k)` = 16(2ⁿ – 1), where the function f satisfies f(x + y) = f(x) . f(y) for all natural numbers x, y and further f(1) = 2.

Answer 1

Given that f(x + y) = f(x) . f(y) and f(1) = 2

Therefore, f(2) = f(1 + 1) = f(1) . f(1) = 2²

f(3) = f(1 + 2) = f(1) . f(2) = 2³

f(4) = f(1 + 3) = f(1) . f(3) = 2⁴

And so on. Continuing the process, we obtain

f(k) = 2^k and f(a) = 2^a

Hence `sum_(k = 1)^n f(a + k) = sum_(k = 1)^n f(a) * f(k)`

= `f(a) sum_(k = 1)^n f(k)`

= 2^a (2¹ + 2² + 2³ + ... + 2ⁿ)

= `2^n {(2*(2^n - 1))/(2 - 1)}`

= `2^(n + 1) (2^n - 1)`  ....(1)

But, we are given `sum_(k = 1)^n f(a + k)` = 16(2ⁿ – 1)

`""^(2^(n + 1)) (2^n - 1)` = 16(2ⁿ – 1)

⇒ 2^a+1 = 2⁴ 

⇒ a + 1 = 4

⇒ a = 3