The following theorem is taken from Levi Ben Gershon's manuscript Maaseh Hoshev (The Art of Calculation), 1321.

When you add consecutive numbers starting with 1, and the number of numbers you add is odd, the result is equal to the product of the middle number among them times the last number.

$$\sum_{i=1}^{2k+1} i = (k+1)(2k+1)$$

Below is my proof by induction to prove this theorem.

Base case (where $$k=0$$):

\begin{align} \sum_{i=1}^{2(0)+1} i &= (0+1)(2(0)+1) \\ \sum_{i=1}^{1} i &= 1\times1 \\ 1 &= 1 \\ \end{align}

So base case checks. Assume $$k = n$$ holds:

$$\sum_{i=1}^{2n+1} i = (n+1)(2n+1)$$

Show $$k = n+1$$ holds:

$$\sum_{i=1}^{2(n+1)+1} i = ((n+1)+1)(2(n+1)+1)$$

\begin{align} \sum_{i=1}^{2(n+1)+1} i &= \\ \sum_{i=1}^{2n+3} i &= \\ \sum_{i=1}^{2n+1} i + \sum_{i=2n+2}^{2n+3} i &= \\ \sum_{i=1}^{2n+1} i + (2n+2) + (2n+3) &= \\ (n+1)(2n+1) + (4n+5) &= \\ (2n^2+3n+1) + (4n+5) &= \\ 2n^2+7n+6 &= \\ (n+2)(2n+3) &= \\ ((n+1)+1)(2(n+1)+1) &= ((n+1)+1)(2(n+1)+1) \\ \end{align}

QED.