\( T_{n}=2 n-3 \) is the general term of the \( 1^{\text {st }} \) level difference of a quadratic sequence and the \( 2^{\text {nd }} \) term is 5 . What is the general term of the quadratic sequence?

**Step 1. Establish the relationship between the quadratic sequence and its first-level differences.** Let the quadratic sequence be given by \[ a_n \] with the first-level differences defined as \[ T_n = a_{n+1} - a_n. \] We are given \[ T_n = 2n - 3. \] **Step 2. Express \(a_n\) in terms of the summation of the differences.** The \(n^{\text{th}}\) term of the quadratic sequence can be expressed as \[ a_n = a_1 + \sum_{k=1}^{n-1} T_k. \] Substitute the expression for \(T_k\): \[ a_n = a_1 + \sum_{k=1}^{n-1} (2k - 3). \] **Step 3. Compute the summation.** Separate the sum: \[ \sum_{k=1}^{n-1} (2k - 3) = 2\sum_{k=1}^{n-1} k - 3\sum_{k=1}^{n-1} 1. \] We know that \[ \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2} \] and \[ \sum_{k=1}^{n-1} 1 = n - 1. \] Substitute these into the expression: \[ \sum_{k=1}^{n-1} (2k - 3) = 2 \cdot \frac{(n-1)n}{2} - 3(n-1) = n(n-1) - 3(n-1). \] Factor out \((n-1)\): \[ \sum_{k=1}^{n-1} (2k - 3) = (n-1)(n - 3). \] Thus, the expression for \(a_n\) becomes: \[ a_n = a_1 + (n-1)(n-3). \] **Step 4. Use the given term of the quadratic sequence to find \(a_1\).** We are given that the second term of the quadratic sequence is \[ a_2 = 5. \] Using our formula: \[ a_2 = a_1 + (2-1)(2-3) = a_1 + 1 \cdot (-1) = a_1 - 1. \] Setting this equal to 5: \[ a_1 - 1 = 5 \quad \Longrightarrow \quad a_1 = 6. \] **Step 5. Write the general term of the quadratic sequence.** Substitute \(a_1 = 6\) back into the expression for \(a_n\): \[ a_n = 6 + (n-1)(n-3). \] Expand \((n-1)(n-3)\): \[ (n-1)(n-3) = n^2 - 3n - n + 3 = n^2 - 4n + 3. \] Thus, \[ a_n = 6 + n^2 - 4n + 3. \] Combine the constant terms: \[ a_n = n^2 - 4n + 9. \] **Final Answer:** \[ a_n = n^2 - 4n + 9. \]