# Closed form expression of a summation

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

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

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

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

I assume the second line should have brackets around the $(3i+1)$. Then:
$$\sum_{i=0}^k(3i+1)=\sum_{i=0}^k3i+\sum_{i=0}^k1$$
$$$$\sum_{i=0}^{k}1 = k+1$$$$
You have an error in the sum : $\sum_0^k 1 =k+1$ and not $1$