Question

Which recursive definition could be used to generate the sequence {3, 6, 3, 6, 3,...}? A) a1 = 3 and an = an-1 + 3n

B) a1 = 3 and an = an-1 + (-3)n
C) a1 = 3 and an = an-1 + 3(-1)n
D) a1 = 3 and an = an-1 + 3(-1)n + 1

Answer 1

The answer is choice C

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You start off with the first term being a1 = 3

Then you add on 3 to get to 6 as the second term

The next thing to do is add on -3, which is the same as subtracting 3, to get back to 3 once more.

This pattern continues on forever. To ensure you add on either +3 or -3 to the previous term, we add on the general term 3*(-1)^n

Overall, the nth term is found by a(n) = a(n-1) + 3*(-1)^n, which is what choice C shows. I'm assuming your teacher meant to write 3(-1)^n instead of 3(-1)n