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I need to give the recursive function of $3n^2$. I'm pretty sure the base case needs to be $3 \cdot 0^2 = 0$, but I don't know where to go from there.

Any help is appreciated.

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1 Answer 1

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If $a_n=3n^2$, then $$a_n-a_{n-1}=3n^2-3(n-1)^2=6n-3$$ Therefore, $$a_n=6n-3+a_{n-1}$$ $$a_0=3\cdot 0^2=0$$ is the recursive relation.

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  • $\begingroup$ Is this the same as a recursive definition? My text book talks about first creating a base case and then a second part that eventually will use the base case. Maybe that doesn't apply to this problem though. $\endgroup$
    – Josh
    Dec 3, 2015 at 1:26
  • $\begingroup$ @Josh Yes you need a base case, since otherwise the recursive relation will keep going down to $-\infty$. You correctly stated in your question that $a_0=0$. $\endgroup$ Dec 3, 2015 at 1:27
  • $\begingroup$ Ah okay, sorry I'm a little confused here. So the entire definition would be my base case and then your relation? $\endgroup$
    – Josh
    Dec 3, 2015 at 1:28
  • $\begingroup$ @Josh Indeed, both parts are necessary for a full recursive formula. I updated the answer to clarify this. $\endgroup$ Dec 3, 2015 at 1:29
  • $\begingroup$ Great, I understand now, thanks. $\endgroup$
    – Josh
    Dec 3, 2015 at 1:30

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