To find the sum \( D = 1 + 3 + 3^2 + 3^3 + \cdots + 3^{100} \), we recognize that this is a geometric series. A geometric series has the form:

\[ S = a + ar + ar^2 + ar^3 + \cdots + ar^n \]

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

In this case:
- The first term \( a = 1 \)
- The common ratio \( r = 3 \)
- The last exponent is 100, so there are \( 101 \) terms (from \( 3^0 \) to \( 3^{100} \)).

The sum \( S \) of the first \( n+1 \) terms of a geometric series can be calculated using the formula:

\[ S = \frac{a(r^{n+1} - 1)}{r - 1} \]

For our series:
- \( a = 1 \)
- \( r = 3 \)
- \( n = 100 \)

So, the sum \( D \) is:

\[ D = \frac{1 \cdot (3^{101} - 1)}{3 - 1} \]
\[ D = \frac{3^{101} - 1}{2} \]

Therefore, the sum of the series \( 1 + 3 + 3^2 + 3^3 + \cdots + 3^{100} \) is:

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