Question

These are the first six terms of a sequence with a1=3 and a2=1: 3, 1, 8, 18, 52, 140, … Find a recursive formula for this sequence that is valid for n>2. Write your answer in the simplest form.

Answer 1

If you add the first two elements (3,1) and multiply the result by 2, you get the third element (8).
In general, to find any element from a3 onwards, you add the previous two and double the result.

As a recursive formula, it will be like:
$a_{1} = 3, a_{2} = 1, a_{n}=a_{n-1} + a_{n-2}$