Final answer:
To solve the recurrence relation s_n = 2s_{n-1} with s_0 = 1, one iterates from the base case to find that the nth term is given by s_n = 2^n.
Step-by-step explanation:
The recurrence relation given is sn = 2sn-1, with the initial condition s0 = 1. This is a simple geometric progression where each term is twice the previous term. To solve the recurrence, we can iterate the relation starting from the base case:
- s1 = 2s0 = 2 × 1 = 2
- s2 = 2s1 = 2 × 2 = 4
- s3 = 2s2 = 2 × 4 = 8
Continuing this process, we can see that the nth term is given by sn = 2n.
This sequence is an example of exponential growth, commonly found in contexts like population dynamics (Nn = N02n) and light refraction (Snell's law), albeit unrelated to the recurrence relation itself.