What are the first six terms of the sequence defined by the recurrence relation a_n = 6a_{n-1} + 4a_{n-2}?

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To find the first six terms of the sequence defined by the recurrence relation an = 6an-1 + 4an-2, we first need the initial conditions. Let’s assume:

  • a0 = 0
  • a1 = 1

Now we can calculate the subsequent terms up to a5:

  1. Term 0: a0 = 0
  2. Term 1: a1 = 1
  3. Term 2: a2 = 6a1 + 4a0 = 6(1) + 4(0) = 6
  4. Term 3: a3 = 6a2 + 4a1 = 6(6) + 4(1) = 36 + 4 = 40
  5. Term 4: a4 = 6a3 + 4a2 = 6(40) + 4(6) = 240 + 24 = 264
  6. Term 5: a5 = 6a4 + 4a3 = 6(264) + 4(40) = 1584 + 160 = 1744

Putting it all together, the first six terms of the sequence are:

  • a0 = 0
  • a1 = 1
  • a2 = 6
  • a3 = 40
  • a4 = 264
  • a5 = 1744

The first six terms are 0, 1, 6, 40, 264, 1744.

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