Demonstrating the speed advantage of parallel computation
Currently each future processes SIZE elements (increasing the NUM_OF_THREADS increases the total workload (NUM_OF_THREADS * SIZE)). Instead you want to divide the workload between each of the futures:

int start = SIZE / NUM_OF_THREADS * iThreadLocal;
int end = SIZE / NUM_OF_THREADS * (iThreadLocal + 1);


Besides that iThread could (and should) be a local variable. The spacing between operators is inconsistent, you should add whitespace around the operators (except for unary operators) to improve readability.