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My prof started out with the following summation: \begin{equation} \sum_{i=0}^{k}i = \frac{k(k+1)}{2} \end{equation} Which is all fine and dandy, however the summation we want to find the closed form of is \begin{equation} \sum_{i=0}^{k}(3i+1) = ? \end{equation} Which I did the following to: \begin{equation} \sum_{i=0}^{k}(3i+1) = 3\frac{k(k+1)}{2} + 1 \end{equation}

However her answer was: \begin{equation} \sum_{i=0}^{k}(3i+1) = 3\frac{k(k+1)}{2} + (k+1) \end{equation}

Is there some summation rule I am missing? How did that $(k+1)$ get added towards the end?

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  • $\begingroup$ Is the second equation $\sum (3i+1)$ or $\left(\sum 3i\right)+1$? $\endgroup$ – Ruvi Lecamwasam Nov 24 '15 at 21:39
  • $\begingroup$ You forgot to sum $1$ (in your $3i+1$) for $i$ from $0$ to $k$, which yields you the term $(k+1)$. $\endgroup$ – TZakrevskiy Nov 24 '15 at 21:39
  • $\begingroup$ Sorry its ∑(3i+1) $\endgroup$ – D. Johnson Nov 24 '15 at 22:17
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I assume the second line should have brackets around the $(3i+1)$. Then:

$$\sum_{i=0}^k(3i+1)=\sum_{i=0}^k3i+\sum_{i=0}^k1$$

As summation is linear, we can apply your prof's formula to the first term.

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$$\begin{equation} \sum_{i=0}^{k}1 = k+1 \end{equation}$$

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You have an error in the sum : $\sum_0^k 1 =k+1$ and not $1$

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