To solve these problems, we need to recognize the patterns in the sequences.

### a) \( A = 1 + 3 + 3^2 + 3^3 + \ldots + 3^{99} + 3^{100} \)

This is a geometric series with the first term \( a = 1 \) and the common ratio \( r = 3 \). The sum of the first \( n \) terms of a geometric series is given by:

\[ S_n = a \frac{r^n - 1}{r - 1} \]

Here, \( n = 101 \) (since the series goes from \( 3^0 \) to \( 3^{100} \)):

\[ A = 1 \frac{3^{101} - 1}{3 - 1} \]
\[ A = \frac{3^{101} - 1}{2} \]

### b) \( B = 2^{100} - 2^{99} + 2^{98} - 2^{97} + \ldots - 2^3 + 2^2 - 2 + 1 \)

This is an alternating series. To simplify, we can group the terms in pairs:

\[ B = (2^{100} - 2^{99}) + (2^{98} - 2^{97}) + \ldots + (2^2 - 2) + 1 \]

Each pair \( (2^k - 2^{k-1}) \) can be simplified as:

\[ 2^k - 2^{k-1} = 2^{k-1}(2 - 1) = 2^{k-1} \]

So, the series becomes:

\[ B = 2^{99} + 2^{97} + \ldots + 2^1 + 1 \]

This is another geometric series with the first term \( a = 2^1 = 2 \) and the common ratio \( r = 2 \). The number of terms is \( 50 \) (since the series goes from \( 2^{99} \) to \( 2^1 \)):

\[ S_n = a \frac{r^n - 1}{r - 1} \]

Here, \( n = 50 \):

\[ B = 2 \frac{2^{50} - 1}{2 - 1} \]
\[ B = 2 (2^{50} - 1) \]
\[ B = 2^{51} - 2 \]

So, the final answers are:

\[ A = \frac{3^{101} - 1}{2} \]
\[ B = 2^{51} - 2 \]