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

To prove that \( \sqrt{S_{2n} - 2S_n} = 2n \), we first need to find the formula for the sum of the first \( n \) terms of an arithmetic sequence with a constant difference of 4. The \( n \)-th term of the sequence can be expressed as: \[ a_n = a_1 + (n-1)d \] where \( d = 4 \) (the common difference). The sum of the first \( n \) terms \( S_n \) is given by: \[ S_n = \frac{n}{2} (2a_1 + (n-1)d) \] Substituting \( d = 4 \): \[ S_n = \frac{n}{2} (2a_1 + (n-1)4) = \frac{n}{2} (2a_1 + 4n - 4) = n(a_1 + 2n - 2) \] Next, we find \( S_{2n} \): \[ S_{2n} = \frac{2n}{2} (2a_1 + (2n-1)4) = n(2a_1 + 8n - 4) \] Now, let's compute \( S_{2n} - 2S_n \): \[ S_{2n} - 2S_n = n(2a_1 + 8n - 4) - 2n(a_1 + 2n - 2) \] Simplifying this: \[ = n(2a_1 + 8n - 4 - 2a_1 - 4n + 4) = n(8n - 4n) = n(4n) \] Thus, we have: \[ S_{2n} - 2S_n = 4n^2 \] Now, taking the square root: \[ \sqrt{S_{2n} - 2S_n} = \sqrt{4n^2} = 2n \] Therefore, we have proved that \( \sqrt{S_{2n} - 2S_n} = 2n \).