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I have this recurrence relation I'm supposed to do, and I can't seem to figure out the next step:

$T_{n} = T_{n-1} + 2n^2, T_{1} = 1$

What I have so far is this:

$T_{n} = T_{n-1} + 2n^2$

$T_{n-2} + 2(n-1) + 2n^2$ <--- Is this right?

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    $\begingroup$ Well... No. $$T_n=T_{n-2}+2(n-1)^2+2n^2$$ $\endgroup$ – user228113 Dec 8 '15 at 3:23
  • $\begingroup$ Where did the $2(n-1)^2$ come from? I'm talking about the exponent. $\endgroup$ – Onikouzou Dec 8 '15 at 3:30
  • $\begingroup$ By the definition $T_s=T_{s-1}+2s^2$ holds for every $s$. By imposing $s=n-1$ you get $T_{n-1}=T_{n-2}+2(n-1)^2$. And you can substitute it in $$T_n=T_{n-1}+2n^2=(T_{n-2}+2(n-1)^2)+2n^2$$ $\endgroup$ – user228113 Dec 8 '15 at 3:34
  • $\begingroup$ Sounds good, thanks. After that I got $\endgroup$ – Onikouzou Dec 8 '15 at 3:39
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    $\begingroup$ You can rewrite it as $T_n - T_{n - 1} = 2n^2$ and now you've got the definition of a summation. So, $T_n = \sum2n^2 = n(2n + 1)(n + 1) / 3 + C$ where $C$ is $-1$. $\endgroup$ – Kaynex Dec 8 '15 at 3:39

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