Suppose $n$ is an integer $\geq 2$. Determine the first digit after the decimal point in the decimal expansion of the number \[\sqrt[3]{n^{3}+2n^{2}+n}\]
1. We start with the given expression \( \sqrt[3]{n^3 + 2n^2 + n} \) and need to determine the first digit after the decimal point in its decimal expansion for \( n \geq 2 \).

2. Consider the bounds for \( n \geq 2 \):
   \[
   (n + 0.6)^3 < n^3 + 2n^2 + n < (n + 0.7)^3
   \]

3. We need to expand both sides to verify these inequalities. First, expand \( (n + 0.6)^3 \):
   \[
   (n + 0.6)^3 = n^3 + 3n^2 \cdot 0.6 + 3n \cdot (0.6)^2 + (0.6)^3
   \]
   \[
   = n^3 + 1.8n^2 + 1.08n + 0.216
   \]

4. Next, expand \( (n + 0.7)^3 \):
   \[
   (n + 0.7)^3 = n^3 + 3n^2 \cdot 0.7 + 3n \cdot (0.7)^2 + (0.7)^3
   \]
   \[
   = n^3 + 2.1n^2 + 1.47n + 0.343
   \]

5. Now, compare these expansions with \( n^3 + 2n^2 + n \):
   \[
   n^3 + 1.8n^2 + 1.08n + 0.216 < n^3 + 2n^2 + n < n^3 + 2.1n^2 + 1.47n + 0.343
   \]

6. Simplify the inequalities:
   \[
   1.8n^2 + 1.08n + 0.216 < 2n^2 + n < 2.1n^2 + 1.47n + 0.343
   \]

7. For \( n \geq 2 \), the inequalities hold true. Therefore, we have:
   \[
   n + 0.6 < \sqrt[3]{n^3 + 2n^2 + n} < n + 0.7
   \]

8. This implies that the first digit after the decimal point in the decimal expansion of \( \sqrt[3]{n^3 + 2n^2 + n} \) is 6.

The final answer is \( \boxed{ 6 } \).