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I'm trying to figure out how reversing the sum works. I know it involves changing or switching variables, but I'm stuck on how to do that.

How do I go from $c + \sum\limits_{i=0}^{n-2} d(n - i)$ to $c + \sum\limits_{i=2}^{n} d(i)$?

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$$\begin{align} \sum_{i=0}^{n-2}d(n-i)&=d(n)+d(n-1)+\dots+d(n-(n-2))\\ &=d(n)+d(n-1)+\dots+d(2)\\ &=d(2)+d(3)+\dots+d(n)\\ &=\sum\limits_{i=2}^{n} d(i) \end{align} $$ Then add both sides by $c$.

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    I understand how you arrived at $d(i)$ but I still don't get how the bounds are changed. – 2017-01-25
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    The issue is to write the finite sum. Look, the first term is $d(2)$ and the last term is $d(n)$, right? So, with respect to $d(i)$, naturally the sum starts at $i=2$ and ends with $i=n$. – 2017-01-25
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    Okay, now I get it, thank you! – 2017-01-25
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    Good. You see the check mark at the left side of my answer?You can click it indicating that you accepted my answer. Thank you – 2017-01-25
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    Thanks for the reminder, kuya. – 2017-01-25