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How do I go about calculating the sum of this finite series? Below is the series and the sum:

$$\sum_{i=0}^{n^2-1}i$$

I understand that you have to use the formula: $\sum_{i=1}^{n}i = \frac{n(n+1)}{2}$ and that the answer is: $\frac{n^4-n^2}{2}$. I just don't know the steps to get there. I'm being thrown off by the starting index of $i=0$.

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    $\begingroup$ The first term contributes nothing to the overall sum... $\endgroup$ – user170231 Nov 10 '15 at 22:56
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Start with the formula

$$ \sum_{i=1}^x{i}=\frac{x(x+1)}{2}. $$

Now, set $x=n^2-1$, and it follows that

\begin{align} \sum_{i=1}^{n^2-1}{i}=\frac{(n^2-1)((n^2-1)+1)}{2}=\frac{n^4-n^2}{2} \end{align}

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The first index adds nothing, so you have $$ \sum_{i=0}^{n^2-1}i=\sum_{i=1}^{n^2-1}i=\frac{(n^2-1)n^2}2. $$

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