From the given information, we get...

\(f(1) = 1 - \frac{1}{2}\)
\(f(2) = \frac{1}{2} - \frac{1}{3}\)
\(f(3) = \frac{1}{3} - \frac{1}{4}\)
\(f(4) = \frac{1}{4} - \frac{1}{5}\)
.
.
.
\(f(199) = \frac{1}{199} - \frac{1}{200}\)
\(f(200) = \frac{1}{200} - \frac{1}{201}\)

So, \(f(1) + f(2) + f(3) + f(4) + . . . . f(199) + f(200) = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) . . . . . + (\frac{1}{199} - \frac{1}{200}) + (\frac{1}{200} - \frac{1}{201}) \)

Notice that almost every POSITIVE fraction has an identical NEGATIVE fraction.
So all of those fractions add up to be ZERO
The only two values that don't cancel out are the first and last terms (1 and -1/201)

In other words, the sum \(= 1 - \frac{1}{201} = \frac{201}{201}-\frac{1}{201}=\frac{200}{201}\)

Answer: D