There are 9 terms in the series \( \sum_{k=2 n}^{n^{2}} \mathrm{~T}_{k} \). Calculate the value of \( n \).

To determine the value of \( n \) for the series \( \sum_{k=2n}^{n^2} T_k \) having 9 terms, we first need to analyze the range of \( k \). The sequence runs from \( k = 2n \) to \( k = n^2 \). The number of terms in this series can be calculated using the formula: \[ \text{Number of terms} = k_{\text{last}} - k_{\text{first}} + 1 = n^2 - 2n + 1 \] We want this expression to equal 9: \[ n^2 - 2n + 1 = 9 \] Now, let's simplify this equation: \[ n^2 - 2n + 1 - 9 = 0 \] This simplifies to: \[ n^2 - 2n - 8 = 0 \] Next, we will factor this quadratic equation. We look for two numbers that multiply to \(-8\) and add to \(-2\). These numbers are \(-4\) and \(2\). Thus, we can factor the equation as: \[ (n - 4)(n + 2) = 0 \] Setting each factor to zero gives us the potential solutions: \[ n - 4 = 0 \quad \Rightarrow \quad n = 4 \] \[ n + 2 = 0 \quad \Rightarrow \quad n = -2 \] Since \( n \) has to be a positive integer (as it is typically defined in series), we take \( n = 4 \). To verify, we substitute \( n = 4 \): \[ \text{Number of terms} = 4^2 - 2 \cdot 4 + 1 = 16 - 8 + 1 = 9 \] This confirms that the answer is correct. Thus, the value of \( n \) is \( \boxed{4} \).