The common difference of an arithmetic sequence is 4 . Prove that \( \sqrt{S_{i n}-2 S_{n}}=2 n \).

To prove that \( \sqrt{S_{2n} - 2 S_n} = 2n \) for an arithmetic sequence with a common difference of 4, let's follow these steps: ### 1. Understanding the Notation - **Arithmetic Sequence**: A sequence where each term increases by a constant difference (\( d \)). Here, \( d = 4 \). - **Sum of the First \( n \) Terms (\( S_n \))**: The sum of the sequence from the first term to the \( n \)-th term. - **Sum of the First \( 2n \) Terms (\( S_{2n} \))**: The sum of the sequence from the first term to the \( 2n \)-th term. ### 2. Formula for the Sum of an Arithmetic Sequence The sum of the first \( m \) terms of an arithmetic sequence is given by: \[ S_m = \frac{m}{2} [2a + (m - 1)d] \] where: - \( a \) is the first term. - \( d \) is the common difference. - \( m \) is the number of terms. ### 3. Calculating \( S_n \) and \( S_{2n} \) Given \( d = 4 \): - **Sum of the First \( n \) Terms (\( S_n \))**: \[ S_n = \frac{n}{2} [2a + (n - 1) \times 4] = \frac{n}{2} [2a + 4n - 4] = n(a + 2n - 2) \] - **Sum of the First \( 2n \) Terms (\( S_{2n} \))**: \[ S_{2n} = \frac{2n}{2} [2a + (2n - 1) \times 4] = n[2a + 8n - 4] = n(2a + 8n - 4) \] ### 4. Calculating \( S_{2n} - 2S_n \) \[ S_{2n} - 2S_n = n(2a + 8n - 4) - 2n(a + 2n - 2) \] \[ = 2an + 8n^2 - 4n - 2an - 4n^2 + 4n \] \[ = (2an - 2an) + (8n^2 - 4n^2) + (-4n + 4n) \] \[ = 4n^2 \] ### 5. Taking the Square Root \[ \sqrt{S_{2n} - 2S_n} = \sqrt{4n^2} = 2n \] (Assuming \( n \) is a positive integer) ### 6. Conclusion Thus, we've proved that: \[ \sqrt{S_{2n} - 2S_n} = 2n \] for an arithmetic sequence with a common difference of 4. **Q.E.D.**