Sum of Identical Terms

Theorem

Let $x$ be a number.

Let $n \in \N$ be a natural number such that $n \ge 1$.


Then:

$\ds \sum_{i \mathop = 1}^n x = n x$


This article, or a section of it, needs explaining.
In particular: Why limit this to $n \ge 1$? It also works for zero.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Explain}} from the code.



Proof


This needs considerable tedious hard slog to complete it.
In particular: this could be actually nontrivial; induction on $n$ seems easiest
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



This article is complete as far as it goes, but it could do with expansion.
In particular: generalize to $x$ an element of a vector space, or for that matter, any abelian group
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Expand}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.