We are asked to evaluate the sum:
\[
S = \sum_{n=1}^{2025} i^n (1+i)
\]
The powers of \( i \) cycle every four terms:
\[
i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \text{and then it repeats.}
\]
Thus, the sum can be written as:
\[
S = \sum_{n=1}^{2025} i^n (1+i)
\]
Since the powers of \( i \) repeat every 4 terms, we can calculate the sum of one complete cycle:
\[
S_{\text{cycle}} = i(1+i) + (-1)(1+i) + (-i)(1+i) + 1(1+i)
\]
Simplifying each term:
\[
S_{\text{cycle}} = i + i^2 + (-i) + i + 1 + i = 2i + 1
\]
Since there are 2025 terms, and 2025 is divisible by 4 (2025 = 4 * 506 + 1), we can split the sum into complete cycles plus the last term:
\[
S = 506 \times (2i + 1) + (i(1+i)) = 506 \times (2i + 1) + i + i^2 = 506 \times (2i + 1) + i - 1
\]
Finally, the sum simplifies to:
\[
S = 2025(1+i)
\]
Thus, the correct answer is:
\[
\boxed{(B) 2025(1+i)}
\]