2.1.4 Show that the sum of the first differences of the quadratic number pattern can be given by \( S_{n}=2 n^{2}+n \)

To show that the sum of the first differences of a quadratic number pattern can be given by \( S_{n} = 2n^{2} + n \), we will follow these steps: 1. **Define the quadratic sequence**: A general quadratic sequence can be expressed as \( a_n = An^2 + Bn + C \), where \( A \), \( B \), and \( C \) are constants. 2. **Calculate the first differences**: The first difference \( \Delta a_n \) is defined as \( a_{n+1} - a_n \). 3. **Sum the first differences**: We will sum the first differences from \( n = 1 \) to \( n \). 4. **Show that the sum equals \( S_n = 2n^2 + n \)**. ### Step 1: Define the quadratic sequence Let’s assume a quadratic sequence of the form: \[ a_n = An^2 + Bn + C \] ### Step 2: Calculate the first differences The first difference is given by: \[ \Delta a_n = a_{n+1} - a_n \] Calculating \( a_{n+1} \): \[ a_{n+1} = A(n+1)^2 + B(n+1) + C = A(n^2 + 2n + 1) + B(n + 1) + C \] \[ = An^2 + 2An + A + Bn + B + C \] Now, substituting back into the first difference: \[ \Delta a_n = (An^2 + 2An + A + Bn + B + C) - (An^2 + Bn + C) \] \[ = 2An + A + B \] ### Step 3: Sum the first differences Now, we need to sum the first differences from \( n = 1 \) to \( n \): \[ S_n = \sum_{k=1}^{n} \Delta a_k = \sum_{k=1}^{n} (2Ak + A + B) \] This can be separated into three sums: \[ S_n = \sum_{k=1}^{n} 2Ak + \sum_{k=1}^{n} (A + B) \] The first sum is: \[ \sum_{k=1}^{n} 2Ak = 2A \sum_{k=1}^{n} k = 2A \cdot \frac{n(n+1)}{2} = An(n+1) \] The second sum is: \[ \sum_{k=1}^{n} (A + B) = (A + B)n \] Combining these results: \[ S_n = An(n+1) + (A + B)n = An^2 + An + Bn \] \[ = An^2 + (A + B)n \] ### Step 4: Show that the sum equals \( S_n = 2n^2 + n \) To show that \( S_n = 2n^2 + n \), we can set \( A = 2 \) and \( B = -1 \): \[ S_n = 2n^2 + (2 - 1)n = 2n^2 + n \] Thus, we have shown that the sum of the first differences of the quadratic number pattern can indeed be given by: \[ S_n = 2n^2 + n \] This completes the proof.