Summation Notation

\sum_{i = m}^{n} x_{i} = x_{m} + x_{m + 1} + \dots + x_{n - 1} + x_{n}

\sum_{i = 0}^{n} x i = x \sum_{i = 0}^{n} i

\sum x \pm \sum y = \sum x \pm y

\sum_{i = m}^{n} x_{i} = \sum_{i = m + k}^{n + k} (x_{i - k})

\sum_{x = 1}^{n} (n - x) = \sum_{x = 1}^{n - 1} x

\begin{aligned} \sum_{i = 1}^{n} (n - i) & = \sum n - \sum i \\ = n \cdot n - \frac{n (n + 1)}{2} \\ = n^{2} - \frac{n^{2} + n}{2} \\ = \frac{2 n^{2} - (n^{2} + n)}{2} \\ = \frac{n^{2} - n}{2} \\ = \frac{n (n - 1)}{2} \\ \sum_{x = 1}^{n} (n - x) & = \sum_{x = 1}^{n - 1} x \neq \sum_{x = 1}^{n} x \end{aligned}

\begin{aligned} \sum_{i = 0}^{n} n - i & = (n + 1) n - \frac{n (n + 1)}{2} \\ = \frac{2 n (n + 1) - n (n + 1)}{2} \\ = \frac{n (n + 1)}{2} \end{aligned}

\sum_{x = 1}^{n} (n - x) = n - n + \sum_{x = 1}^{n - 1} (n - x) = \sum_{x = 1}^{n - 1} (n - x)

\sum_{i = 1}^{n} i^{2} = \frac{n \cdot (n + 1) \cdot (2 n + 1)}{6}

\sum_{i = 1}^{n} i^{3} = {(\sum_{i = 1}^{n} i)}^{2} = {(\frac{n (n + 1)}{2})}^{2}

\sum_{i = 0}^{n} a^{i} = \frac{a^{n + 1} - 1}{a - 1}