D = 1 - 2 + 2^2 - 2^3 + 2^4 - .... - 2^33

To solve for \( D \) in the series

\[
D = 1 - 2 + 2^2 - 2^3 + 2^4 - \cdots - 2^{33}
\]

we can first rewrite the series more clearly:

\[
D = \sum_{n=0}^{33} (-2)^n
\]

This is a finite geometric series with the first term \( a = 1 \) (when \( n=0 \)) and the common ratio \( r = -2 \). The number of terms in the series is \( 34 \) (from \( n=0 \) to \( n=33 \)).

The formula for the sum of a geometric series is given by:

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

where \( S_n \) is the sum of the first \( n \) terms, \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

For our series, we have:

- \( a = 1 \)
- \( r = -2 \)
- \( n = 34 \)

Now we substitute these values into the formula:

\[
D = 1 \cdot \frac{1 - (-2)^{34}}{1 - (-2)}
\]

Calculating the denominator:

\[
1 - (-2) = 1 + 2 = 3
\]

Calculating the numerator where \( (-2)^{34} = 2^{34} \):

\[
1 - (-2)^{34} = 1 - 2^{34}
\]

Now substituting back gives:

\[
D = \frac{1 - 2^{34}}{3}
\]

Thus,

\[
D = \frac{1 - 2^{34}}{3}
\]

This is our final answer:

\[
\boxed{\frac{1 - 2^{34}}{3}}
\]