\((1 / 2)+(3 / 4)+(7 / 8)+(15 / 16) \ldots 100\) terms \(=\) (A) \(2^{100}+99\) (B) \(2^{-100}+99\) (C) \(2^{-101}+100\) (D) \(2^{-99}+99\)

Before finding the sum of the series, let's look at the pattern of the given series. The series is: \[ \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} \ldots \text{[100 terms]} \] Notice that the series can be written as: \[ \frac{2^1 - 1}{2^1} + \frac{2^2 - 1}{2^2} + \frac{2^3 - 1}{2^3} + \frac{2^4 - 1}{2^4} \ldots \]

Observe that the numerators have a power of 2 minus 1 pattern, so when you add consecutive terms, there'll be a cancellation effect taking place. Let's see what happens when we add the first term to the second term and so on: \[ \left(1 - \frac{1}{2}\right) + \left(1 - \frac{1}{4}\right) + \left(1 - \frac{1}{8}\right) + \left(1 - \frac{1}{16}\right) \ldots \]

From the pattern above, we can add the consecutive terms so that the series becomes the sum of the \(100\) terms minus the sum of the fractions \(1/2, 1/4, 1/8, ...\) up to the \(99th\) term. Now let's find the sum of the fractions, which is a geometric series with the common ratio of \(1/2\). The sum of an infinite geometric series is given by the formula \(\frac{a}{1 - r}\), where \(a\) is the first term and \(r\) is the common ratio. However, since we want the sum of the first 99 terms, the formula will be \(\frac{a(1 - r^n)}{1 - r}\), where n is the number of terms. Sum of fractions: \[ S = \frac{\frac{1}{2}(1 - \left(\frac{1}{2}\right)^{99})}{1 - \frac{1}{2}} = 1 - \left(\frac{1}{2}\right)^{99} \] So, the sum of the given series is: Sum of the given series = \(100\) - Sum of fractions \[ 100 - ( 1 - \left(\frac{1}{2}\right)^{99} ) = \left(\frac{1}{2}\right)^{99} + 99 \] Thus, the sum of the series is: \[ \left(\frac{1}{2}\right)^{99} + 99 \] Comparing the result to the given options, the sum of the series matches with option (D). Answer: \(\boxed{(D)}\).