Question

Let the function \(f: \mathrm{N} \rightarrow \mathrm{R}\) satisfy \(f(x+y)=f(x) \cdot f(y)\) for all \(x, y \in \mathrm{N}\) and \(f(1)=2\). If \(\sum_{\mathrm{k}=1}^{50} f(4+\mathrm{k})=\lambda\left(2^{\beta}-1\right)\), then \(\lambda+\beta\) is equal to ________.

Hint:

Show f(n) = 2ⁿ, so the sum is a geometric series with first term 2⁵ and ratio 2.

Answer 1

Correct Answer: 82

f(n + 1) = f(n) f(1) = 2 f(n), so f(n) = 2ⁿ for all natural n.
Sum of f(4 + k) for k = 1 to 50 = 2⁵ + 2⁶ + ... + 2⁵⁴, a G.P. with 50 terms, first term 32 and ratio 2.
Sum = 32 (2⁵⁰ - 1)/(2 - 1) = 32 (2⁵⁰ - 1).
Comparing with λ(2^β - 1): λ = 32 and β = 50, so λ + β = 82.
Answer: 82