by showmyiq » Wed Sep 26, 2012 10:10 am
Because we are going to use that very often, I would like to write down simple and beautiful proof.
We need to calculate the sum of all natural numbers from 1 to n.
\begin{equation}
S_n = 1 + 2 + 3 + .... + n
\end{equation}
Let’s arrange it like that:
Sn = 1 + 2 + 3 + … + (n-1) + n
If we multiply the equation both sides by 2, we get this:
[1 + 2 + 3 + … + (n-1) + n] + [1 + 2 + 3 + … + (n-1) + n] = 2*Sn
Let’s rearrange the left side like that
[1 + 2 + 3 + … + (n-1) + n] +
[n + (n-1) + ….+ 2 +1] =
=2*Sn
Do you see the pattern now? If we add the first parts of the elements in the left side of the equation, we get n+1. If we add the second parts – 2 and (n-1), we again get n+1. If we continue like that till the end elements, we again receive n+1. We have a total of n pairs, so the result of the left side of the equation will be n pairs multiplied by (n+1), which is their value. So we get
n*(n+1) = 2*Sn
So at the end we receive:
\begin{equation}
S_n =\frac {n*(n+1)} {2}
\end{equation}
You can share your solutions below!