## martes, 27 de noviembre de 2012

### A mathematical induction exercise

Nowadays, I am taking some classes of abstract algebra at UNED, because mathematics is a powerful tool to research and I have in mind starting Ph.D studies (I have a master degree) in the future (perhaps). I want to talk about mathematical induction in this post for practicing Maths and improving my bad English too.
We have to demonstrate that if $$n\geq 1$$, then: $$1+2+3+...+n=\frac{n(n+1)}{2}$$ Happens for any number. First we check that this happens for 1. $$1 = \frac{1(1+1)}{2} = 1$$ And, we need to check for $$n+1$$. $$1+2+3+...+ n+(n+1)=\frac{(n+1)((n+1)+1)}{2}=$$ $$\frac{(n+1)^{2}+(n+1)}{2}=\frac{n^{2}+2n+1+n+1}{2}=$$ $$\frac{n^{2}+n+2n+2}{2}=\frac{n(n+1)+2n+2}{2}=\frac{n(n+1)}{2}+(n+1)$$ Therefore, as we can see: $$1+2+3+...+ n + (n + 1) = \frac{n(n+1)}{2}+(n+1)$$
Quod erat demonstrandum