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Give recursive definition of sequence $a_n = 2^n, n = 2, 3, 4... where $ $a_1 = 2$

I'm just not sure how to approach these problems.

Then it asks to give a def for:

$a_n = n^2-3n, n = 0, 1, 2...$

Thanks for all the help!

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  • $\begingroup$ The first one is easy: $a_{n+1}=2a_n$. For the next one, can you similarly solve for $a_{n+1}$ in terms of the previous terms? $\endgroup$ Aug 4, 2015 at 0:15
  • $\begingroup$ So is that the definition (the answer then). Could you just briefly explain how to get there? I'm new to these. Thanks so much! $\endgroup$
    – androidguy
    Aug 4, 2015 at 0:16

2 Answers 2

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For the first one, write the term $a_{n+1}$ and compare it to $a_n$: $$a_{n+1}=2^{n+1}=2\cdot2^n=2a_n$$

For the second one, repeat the process: $$\begin{align}a_{n+1}&=(n+1)^2-3(n+1)\\ &=n^2+2n+1-3n-3\\ &=n^2-3n+2n-2\\ &=a_n+2(n-1) \end{align}$$

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  • $\begingroup$ Okay, great! Thanks for showing how to do it so I can figure out future problems. Really appreciate the help! $\endgroup$
    – androidguy
    Aug 4, 2015 at 0:26
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Note that there are lots of answers to these questions. IMO those given by Terra Hyde are probably what your instructor is expecting, but you could also say:

for the first one, $$a_{n+1}=a_n+2^n$$ and for the second $$a_{n+1}=a_n+a_{n-1}-n^2+7n-6$$ among many, many other possibilities.

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  • $\begingroup$ Yeah, it makes sense that there would be multiple answers for some of these. Thanks for letting me know, appreciate the help! $\endgroup$
    – androidguy
    Aug 4, 2015 at 0:30
  • $\begingroup$ I had sort of a follow up question for anyone here, in another problem, I'm asked to give a recursive def of the set of positive integers that are a multiple of 4. So is this sort of the same thing, just working backwards a bit? $\endgroup$
    – androidguy
    Aug 4, 2015 at 0:33
  • $\begingroup$ Recursive definition of a set is a bit different. Suggest you post that as a separate question. $\endgroup$
    – David
    Aug 4, 2015 at 0:39

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