Brain works 2

$\quad \; f(x)=1-f(1-x) \\ \rightarrow f(x)+f(1-x)=1$

The form of the function tells us that, Gaussian pairing tool might be useful here.

Let, $S = f(\frac{1}{999})+f(\frac{2}{999})+f(\frac{3}{999})+...+f(\frac{997}{999})+f(\frac{998}{999})$ $\ \\ \rightarrow S = f(\frac{998}{999})+f(\frac{997}{999})+f(\frac{996}{999})+...+f(\frac{2}{999})+f(\frac{1}{999})$

With addition, $2S = (f(\frac{1}{999})+f(\frac{998}{999})) + (f(\frac{2}{999})+f(\frac{997}{999})) + ... + (f(\frac{998}{999})+f(\frac{1}{999})) \\ . \\ \quad = \sum_{a=1}^{998} (f(a)+f(1-a)) \\ \quad = \sum_{a=1}^{998} 1 = 998$

Thus we get, $2S=998 \rightarrow S=\boxed{499}$ which is the desired answer.

a(n) = a + d(n-1)

$\frac{998}{999}$ = $\frac{1}{999}$ + $\frac{1}{999}$ (n-1), n = 998, which can easily be seen has 998 terms as the function is the complement of the complement or itself. ex. f( $\frac{1}{999}$ ) = $\frac{1}{999}$

S = $\frac{998}{2}$ (2( $\frac{1}{999}$ )+(998-1) $\frac{1}{999}$ )

S = 499

• $\frac{998x999}{2}$ = $\frac{498501}{999}$ =499