The sum of the digits in base ten of $ (10^{4n^2\plus{}8}\plus{}1)^2$, where $ n$ is a positive integer, is

$ \textbf{(A)}\ 4 \qquad 
\textbf{(B)}\ 4n \qquad 
\textbf{(C)}\ 2\plus{}2n \qquad 
\textbf{(D)}\ 4n^2 \qquad 
\textbf{(E)}\ n^2\plus{}n\plus{}2$
1. Let's start by examining the expression \((10^{4n^2 + 8} + 1)^2\). We can rewrite it as:
   \[
   (10^X + 1)^2 \quad \text{where} \quad X = 4n^2 + 8
   \]
2. Expanding the square, we get:
   \[
   (10^X + 1)^2 = 10^{2X} + 2 \cdot 10^X + 1
   \]
3. Substituting \(X = 4n^2 + 8\), we have:
   \[
   10^{2(4n^2 + 8)} + 2 \cdot 10^{4n^2 + 8} + 1
   \]
4. Notice that \(10^{2(4n^2 + 8)}\) is a 1 followed by \(2(4n^2 + 8)\) zeros, \(2 \cdot 10^{4n^2 + 8}\) is a 2 followed by \(4n^2 + 8\) zeros, and the final 1 is just 1. When we add these together, the result is:
   \[
   1\underbrace{00\ldots0}_{4n^2 + 8 \text{ zeros}}2\underbrace{00\ldots0}_{4n^2 + 8 \text{ zeros}}1
   \]
5. The sum of the digits of this number is:
   \[
   1 + 2 + 1 = 4
   \]
6. Therefore, the sum of the digits in base ten of \((10^{4n^2 + 8} + 1)^2\) is 4.

The final answer is \(\boxed{4}\)