Then, $\sum _{i=1} ^{k+1} i = \biggl(\sum _{i=1}^k i \biggl) + (k+1)$
$=\frac{(k+\frac{1}{2})^2}{2}+k+1 \quad (\text{by inductive hypothesis}).$
$=\frac{k^2+k+\frac{1}{4}}{2} + k +1$$=\frac{k^2+3k+\frac{9}{4}}{2} =\frac{(k+\frac{3}{2})^2}{2} $
$=\frac{[(k+1) + \frac{1}{2}]^2}{2}, \ \ \text{completing the inductive step}$
Hence, by mathematical induction, for every positive integer $n$
$\sum_{i=1}^n i = \frac{(n+\frac{1}{2})^2}{2} $