The solution to the recurrence relation an=3a(n-1)-  a(n-2), where a0=1 and a1=3, then an= ________.Question 26Answera.(1-n)b.(n-2)c.(1+n)d.2n ⋅

To solve the recurrence relation an = 3a(n-1) - a(n-2), where a0 = 1 and a1 = 3, we need to find a pattern in the sequence.

Let's start by calculating the first few terms:

a0 = 1 (given) a1 = 3 (given) a2 = 3a1 - a0 = 33 - 1 = 8 a3 = 3a2 - a1 = 38 - 3 = 21 a4 = 3a3 - a2 = 321 - 8 = 55 a5 = 3a4 - a3 = 355 - 21 = 144

Looking at these terms, it's not immediately clear what the pattern is. However, if we look at the differences between consecutive terms, we see:

a1 - a0 = 3 - 1 = 2 a2 - a1 = 8 - 3 = 5 a3 - a2 = 21 - 8 = 13 a4 - a3 = 55 - 21 = 34 a5 - a4 = 144 - 55 = 89

These differences form a sequence that is very similar to the Fibonacci sequence, but shifted by one and multiplied by 2. This suggests that the solution to the recurrence relation is related to the Fibonacci sequence.

However, none of the options provided (a. (1-n), b. (n-2), c. (1+n), d. 2n) seem to match this pattern. It's possible that there's a mistake in the question or the provided options.