Question

Show that sum of '$n$' natural numbers is $\frac{n(n+1)}{2}$

Answer 1

The first n natural numbers will form an AP with the common difference 1, so for the sum we can use the sum of terms in AP formula;

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_n=\frac{n}{2}[2+(n-1\left)1\right]=\frac{n(n+1)}{2}$