Suppose that $n$ is s positive integer. Determine all the possible values of the first digit after the decimal point in the decimal expression of the number $\sqrt{n^3+2n^2+n}$
To determine the first digit after the decimal point in the decimal expression of the number \(\sqrt{n^3 + 2n^2 + n}\), we can follow these steps:

1. **Simplify the expression inside the square root:**
   \[
   \sqrt{n^3 + 2n^2 + n} = \sqrt{n(n^2 + 2n + 1)} = \sqrt{n(n+1)^2} = (n+1)\sqrt{n}
   \]

2. **Express the first digit after the decimal point:**
   The first digit after the decimal point of \((n+1)\sqrt{n}\) can be found by considering \(10(n+1)\sqrt{n}\) and taking the floor function:
   \[
   N = \lfloor 10(n+1)\sqrt{n} \rfloor
   \]
   Let \(N = 10k + r\), where \(k\) is an integer and \(0 \leq r < 10\).

3. **Set up the inequality:**
   \[
   10k + r \leq 10(n+1)\sqrt{n} < 10k + r + 1
   \]
   Dividing by 10:
   \[
   k + \frac{r}{10} \leq (n+1)\sqrt{n} < k + \frac{r+1}{10}
   \]

4. **Square the inequality:**
   \[
   \left(k + \frac{r}{10}\right)^2 \leq n(n+1)^2 < \left(k + \frac{r+1}{10}\right)^2
   \]
   Expanding both sides:
   \[
   k^2 + \frac{2kr}{10} + \frac{r^2}{100} \leq n(n+1)^2 < k^2 + \frac{2k(r+1)}{10} + \frac{(r+1)^2}{100}
   \]
   Simplifying the right-hand side:
   \[
   k^2 + \frac{2kr}{10} + \frac{r^2}{100} \leq n(n+1)^2 < k^2 + \frac{2kr}{10} + \frac{2k}{10} + \frac{r^2}{100} + \frac{2r}{100} + \frac{1}{100}
   \]
   \[
   k^2 + \frac{2kr}{10} + \frac{r^2}{100} \leq n(n+1)^2 < k^2 + \frac{2kr}{10} + \frac{2k}{10} + \frac{r^2}{100} + \frac{2r}{100} + \frac{1}{100}
   \]

5. **Analyze the inequality:**
   From the inequality, we can see that the value of \(r\) (the first digit after the decimal point) depends on the value of \(n\) and \(k\). By trial and error, we can find that all values of \(r\) from 0 to 9 are achievable by suitable choices of \(n\) and \(k\).

Conclusion:
The first digit after the decimal point in the decimal expression of \(\sqrt{n^3 + 2n^2 + n}\) can be any digit from 0 to 9.

The final answer is \(\boxed{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}\).