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 Σₖ₌₁ⁿ f(k) = 3279, then the value of n is

Solution

Maths sir · 4 steps

Build f(n) one step at a time from f(1) = 3 and you get powers of 3, so the sum is a GP.

1. Find f(n). Put y = 1: f(x+1) = f(x) × f(1) = 3f(x). Each value is 3 times the previous one, so f(2) = 9, f(3) = 27, and in general f(k) = 3ᵏ.
2. Sum the GP. The sum 3 + 9 + 27 + … + 3ⁿ is a GP with first term 3 and common ratio 3.

   Σₖ₌₁ⁿ 3ᵏ = 3(3ⁿ - 1)/(3 - 1) = 3279

3. Solve for n. 3ⁿ - 1 = (2 × 3279)/3 = 2186, so 3ⁿ = 2187 = 3⁷. As a check, 3 + 9 + 27 + 81 + 243 + 729 + 2187 = 3279.
4. So. n = 7, which is option (b).

The idea to keep. Whenever f(x+y) = f(x)f(y) on the naturals, f(n) = (f(1))ⁿ. Any sum of such values is then a GP, so use the GP sum formula.

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