How do you find the first six terms of the sequence defined by the recurrence relation a(n) = 4a(n-1) + 2a(n-2) with initial terms a(0) = 1 and a(1) = 2?

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The first step to finding the first six terms of a sequence defined by a recurrence relation is to identify the recurrence relation itself, initial terms, and to use these to calculate subsequent terms. The recurrence relation given is:

a(n) = 4a(n-1) + 2a(n-2)

We also have the initial terms:

  • a(0) = 1
  • a(1) = 2

Now, we can compute further terms of the sequence step by step:

1. Calculate a(2):

Using the recurrence relation:

a(2) = 4a(1) + 2a(0) = 4(2) + 2(1) = 8 + 2 = 10

2. Calculate a(3):

a(3) = 4a(2) + 2a(1) = 4(10) + 2(2) = 40 + 4 = 44

3. Calculate a(4):

a(4) = 4a(3) + 2a(2) = 4(44) + 2(10) = 176 + 20 = 196

4. Calculate a(5):

a(5) = 4a(4) + 2a(3) = 4(196) + 2(44) = 784 + 88 = 872

Now that we have calculated all necessary terms, we can summarize:

Final Terms of the Sequence:

  • a(0) = 1
  • a(1) = 2
  • a(2) = 10
  • a(3) = 44
  • a(4) = 196
  • a(5) = 872

The first six terms of the sequence are therefore:

1, 2, 10, 44, 196, 872

This step-by-step process of calculating each term leads us to the desired result.

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