Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality

In the setting of Commutative Rings, Fields and Ordered Fields: Standard Notation, let $n\in\mathbb{N}$, with $[n]$ as in Intervals of Natural Numbers §segment.

Let $R$, with $+$, $\cdot$, $0$ and $1$, be a commutative ring; let $a,b:[n]\to R$ and $c\in R$.

¶ $c\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}c\,a_{k}$.

¶ $-\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}(-a_{k})$ and $\sum_{k=1}^{n}(a_{k}-b_{k})=\sum_{k=1}^{n}a_{k}-\sum_{k=1}^{n}b_{k}$.

¶ For $d:[n+1]\to R$, $\sum_{k=1}^{n}(d_{k+1}-d_{k})=d_{n+1}-d_{1}$.

¶ $\sum_{k=1}^{n}c=n\,c$.

Let now $F$ be an ordered field, and let $a,b:[n]\to F$.

¶ If $a_{k}\le b_{k}$ for every $k\in[n]$, then $\sum_{k=1}^{n}a_{k}\le\sum_{k=1}^{n}b_{k}$; if moreover $a_{j}<b_{j}$ for some $j\in[n]$, then $\sum_{k=1}^{n}a_{k}<\sum_{k=1}^{n}b_{k}$.

¶ If $a_{k}\ge0$ for every $k\in[n]$, then $a_{j}\le\sum_{k=1}^{n}a_{k}$ for every $j\in[n]$; in particular $\sum_{k=1}^{n}a_{k}\ge0$.

¶ $\Big|\sum_{k=1}^{n}a_{k}\Big|\le\sum_{k=1}^{n}|a_{k}|$.