Question

For each of the following, solve the given recurrence relation. Show your workings.

(i) 3 an-1 for n > 1, at a1= 7.
(ii) an = 4an-1 + 6 for n > 0, a0 = 1. Hint: describe it as a summation.
(iii) an+1 =1 = 2nan for n > I, a1= 1.

Answer 1

The recurrence relations are solved individually, with the first being a geometric sequence, the second involving iteration or summation for each term, and the third being potentially malformed but interpretable as a rapidly growing sequence where each term is the previous term multiplied by 2n.

Step-by-step explanation:

To solve the given recurrence relations, we need to follow each problem individually:

1. For the first recurrence relation an = 3an-1 with a1 = 7, this is a geometric sequence where each term is three times the previous term. We can solve it by using the formula for the n-th term of a geometric sequence which is a1 * r^(n-1), where r is the common ratio. In this case, r = 3, hence the n-th term an = 7 * 3^(n-1).
2. For the second recurrence relation, an = 4an-1 + 6 with a0 = 1, we can solve this by iterating the formula or describing it in terms of a summation. We start by plugging in values for n to get a sequence of values for an. As n increases, we can express the non-homogeneous part (the constant term) as a sum of 6's that grows with n.
3. The third recurrence relation is slightly malformed, but assuming it is meant to be an+1 = 2n * an with a1 = 1, it suggests that we multiply the previous term by 2n to get the next term. This sequence grows extremely fast since the multiplier doubles each time. We can express the n-th term using the iteration of the given formula starting from the first term.

All these are examples of sequence and series expansions, particularly related to the growth of sequences defined by recurrence relations.