To evaluate the expression

\[
S = 3 + 2^2 - 2^3 + 2^4 - 2^5 + \ldots - 2^{99} + 2^{100},
\]

we can first rewrite \( S \) for clarity:

\[
S = 3 + \sum_{n=2}^{100} (-1)^{n} 2^n.
\]

Notice that the terms \(2^2, -2^3, 2^4, -2^5, \ldots, -2^{99}, 2^{100}\) can be grouped in pairs. Each pair \( (2^n - 2^{n+1}) \) will alternate in sign depending on whether \( n \) is even or odd.

To rewrite this, we can consider the following:

### Step 1: Pairing the Terms
We notice that:

\[
2^2 - 2^3 = 2^2(1 - 2) = -2^2,
\]
\[
2^4 - 2^5 = 2^4(1 - 2) = -2^4,
\]
\[
2^6 - 2^7 = 2^6(1 - 2) = -2^6,
\]
and so forth until \( 2^{98} - 2^{99} = 2^{98}(1 - 2) = -2^{98}.
\]

Thus, we can sum these pairs:

\[
S = 3 + (-2^2 - 2^4 - 2^6 - \ldots - 2^{98}) + 2^{100}.
\]

### Step 2: Summing the Series
The summation \( - (2^2 + 2^4 + 2^6 + \ldots + 2^{98}) \) is a geometric series with the first term \( 2^2 \) and a common ratio of \( 2^2 = 4 \). The number of terms in this series is \( 49 \) (it runs from \( 2 \) to \( 98 \) by increments of \( 2 \)).

We can calculate the sum of the geometric series as follows:

\[
\text{Sum} = a \frac{r^n - 1}{r - 1},
\]
where \( a = 2^2 = 4 \), \( r = 4 \), and \( n = 49 \).

So,

\[
\text{Sum} = 4 \frac{4^{49} - 1}{4 - 1} = \frac{4^{50} - 4}{3}.
\]

### Step 3: Putting it All Together
Now we have:

\[
S = 3 - \left( \frac{4^{50} - 4}{3} \right) + 2^{100}.
\]

We also note that \(2^{100} = 4^{50}\). Therefore:

\[
S = 3 - \left( \frac{4^{50} - 4}{3} \right) + 4^{50} = 3 + \frac{4}{3} + \frac{4^{50} \cdot 3 - 4^{50}}{3} = 3 + \frac{4}{3} + \frac{2 \cdot 4^{50}}{3}.
\]

### Final Calculation
Using common denominator, we get:

\[
S = \frac{9}{3} + \frac{4}{3} + \frac{2 \cdot 4^{50}}{3} = \frac{13 + 2 \cdot 4^{50}}{3}.
\]

Thus, the final answer is

\[
\boxed{\frac{13 + 2 \cdot 4^{50}}{3}}.
\]