The sequence provided alternates between subtracting \(3^{2k}\) and adding \(3^{2k+1}\). Let's analyze the pattern:

\[ D = -1 + 3 - 3^2 + 3^3 - 3^4 + 3^5 - \cdots - 3^{2n} + 3^{2n+1} \]

The terms of the sequence can be grouped into pairs with opposite signs:

For \(k = 0\) to \(k = n\), consider the terms:

\[ (-3^{2k}) + 3^{2k+1} = 3^{2k+1} - 3^{2k} \]

Thus, each pair of terms simplifies to:

\[ 3^{2k} \times (3 - 1) = 3^{2k} \times 2 = 2 \times 3^{2k} \]

This means you can factor out 2 from each complete pair:

\[ D = \sum_{k=0}^{n} (2 \times 3^{2k}) = 2 \sum_{k=0}^{n} 3^{2k} \]

The sum inside is a geometric series:

\[ \sum_{k=0}^{n} 3^{2k} = 1 + 3^2 + (3^2)^2 + \cdots + (3^2)^n \]

This is a geometric series with first term \(a = 1\) and common ratio \(r = 3^2 = 9\). The sum of a geometric series is:

\[ S_n = \frac{a(r^{n+1} - 1)}{r - 1} = \frac{1 \times (9^{n+1} - 1)}{9 - 1} = \frac{9^{n+1} - 1}{8} \]

Thus, the entire expression for \(D\) becomes:

\[ D = 2 \times \frac{9^{n+1} - 1}{8} = \frac{2(9^{n+1} - 1)}{8} \]

\[ D = \frac{9^{n+1} - 1}{4} \]

Therefore, the value of \(D\) is:

\[ D = \frac{9^{n+1} - 1}{4} \]