Circular Sums

$S = \displaystyle \sum_{n=0}^{2016} e^{\dfrac{ni\pi}{2}} = \sum_{n=0}^{2016} (i)^{n} = 1 + (i+i^2+i^3+i^4) + (i^5 + i^6 + i^7 + i^8) + \ldots + (i^{2013} + i^{2014} +i^{2015} + i^{2016} ) = 1 + (0) + (0) + \ldots + (0) = 1$

Comparing $a = 1, b = 0$
$a + b = 1$

Evaluating the first 4 terms would give: $e^{0}$ =1, $e^{i\pi/2}$ =i, $e^{i\pi}$ =-1 and $e^{3i\pi/2}$ =-i

Clearly this means that whenever n is in the form 4m-1 the sum is 0. 4|2016 therefore the sum is 0+ $e^{2016i\pi/2}$ which is 1.