Let f : N → R be a function such that f(x + y) = 2f(x)f(y) for natural numbers x and y. If f(1) = 2, then the value of α for which
Σₖ₌₁¹⁰ f(α + k) = 512/3(2²⁰ - 1)
holds, is :

Solution

Maths sir · 4 steps

Absorb the stray 2 into a new function, and the equation becomes the plain exponential rule.

1. Clean up the 2. Let g(x) = 2f(x). Multiplying the given equation by 2 gives g(x+y) = g(x) g(y), with g(1) = 2 × 2 = 4.
2. Find f(n). A function with g(x+y) = g(x)g(y) on natural numbers is g(n) = g(1)ⁿ = 4ⁿ. So f(n) = 4ⁿ/2. Check: f(2) = 2f(1)² = 8 = 16/2.

   Σₖ₌₁¹⁰ f(α+k) = 4^α/2Σₖ₌₁¹⁰4ᵏ = 4^α/2·4(4¹⁰-1)/3 = (2 · 4^α)/3(2²⁰-1)

3. Match both sides. Here 4¹⁰ = 2²⁰. Comparing with 512/3(2²⁰-1) gives 2 · 4^α = 512, so 4^α = 256 = 4⁴.
4. So. α = 4, which is option (c).

The idea to keep. When a functional equation reads f(x+y) = c f(x)f(y), put g = cf. Then g(x+y) = g(x)g(y), so g(n) = g(1)ⁿ, and the sum becomes a GP.

Answer from the NTA final answer key. Solution by PrepWiser.