Proof Problem - 3

A much simple proof for this question is, let the sum of them be $\zeta$ , then we will get $\zeta = 1+2+3+\dots+n$

Now reverse it, then we get $\zeta = n+(n-1)+(n-2)+\dots+3+2+1$

Combining them, we will get $2\zeta = \underset{n\:terms}{\underbrace{(n+1)+(n+1)+\dots+(n+1)}} = n(n+1)$

Hence $\zeta = \sum_{m=1}^n m = \frac{n(n+1)}{2}$

First, let us see if this is true for $\large n = 1$

$\large \frac {1(1 + 1)}{2} = \frac {1(2)}{2} = \frac {2}{2} = 1$ , $\large 1 = 1$ so it is true for $\large n = 1$

Let's assume it is true for $\large n = r$ , so that $\large 1 + 2 + 3 + \dots + r = \frac {r(r + 1)}{2}$

Now let us see if it is true for $\large n = r + 1$

$\large 1 + 2 + 3 + \dots + r + (r + 1) = \frac {r(r + 1)}{2} + (r + 1)$

That equals $\large \frac {r(r + 1) + 2(r + 1)}{2} = \frac {(r + 1)(r + 2)}{2} = \frac {(r + 1)((r + 1) + 1)}{2}$

If the statement is true for $\large r$ and $\large r + 1$ , it must be true for all positive integers. Q.E.D.