Let's answer both questions step by step as requested: --- ## **Question 2** Given the sequence: \( a_n = n^2 + 3n + 4 \) ### **Step 1: Find the sequence of differences** Compute the difference \( a_n - a_{n-1} \): \[ a_n - a_{n-1} = [n^2 + 3n + 4] - [(n-1)^2 + 3(n-1) + 4] \] Expand \( (n-1)^2 = n^2 - 2n + 1 \): \[ a_{n-1} = (n-1)^2 + 3(n-1) + 4 = (n^2 - 2n + 1) + (3n - 3) + 4 = n^2 - 2n + 1 + 3n - 3 + 4 \] \[ = n^2 + (3n - 2n) + (1 - 3 + 4) \] \[ = n^2 + n + 2 \] So, \[ a_n - a_{n-1} = [n^2 + 3n + 4] - [n^2 + n + 2] = (n^2 - n^2) + (3n - n) + (4 - 2) = 2n + 2 \] **Explanation:** We computed the difference between consecutive terms and found \( a_n - a_{n-1} = 2n + 2 \). This is an arithmetic sequence. --- ### **Step 2: Write the recursion relation** From above: \[ a_n - a_{n-1} = 2n + 2 \implies a_n = a_{n-1} + 2n + 2 \] **Explanation:** We rearranged the difference to get the recursion relation in terms of \( a_{n-1} \). --- **Final Answer for Question 2:** \[ \boxed{a_n = a_{n-1} + 2n + 2} \] where \( a_1 = 1^2 + 3 \times 1 + 4 = 8 \). --- ## **Question 3** Given the general quadratic sequence: \( s_n = a n^2 + b n + c \) ### **Step 1: Find the sequence of differences** Compute the difference \( s_n - s_{n-1} \): \[ s_n - s_{n-1} = [a n^2 + b n + c] - [a(n-1)^2 + b(n-1) + c] \] Expand \( (n-1)^2 = n^2 - 2n + 1 \): \[ a(n-1)^2 = a(n^2 - 2n + 1) = a n^2 - 2a n + a \] \[ b(n-1) = b n - b \] So, \[ s_{n-1} = a n^2 - 2a n + a + b n - b + c \] Subtract: \[ s_n - s_{n-1} = [a n^2 + b n + c] - [a n^2 - 2a n + a + b n - b + c] \] \[ = a n^2 - a n^2 + b n - b n + c - c + 2a n - a + b \] \[ = 2a n + (b - a) \] **Explanation:** We expanded and subtracted to show that the difference is \( 2an + (b - a) \), which is an arithmetic sequence. --- ### **Step 2: Write the recursion relation** \[ s_n - s_{n-1} = 2a n + (b - a) \implies s_n = s_{n-1} + 2a n + (b - a) \] **Explanation:** We rearranged the difference formula into a recursion relation in terms of \( s_{n-1} \). --- **Final Answer for Question 3:** \[ \boxed{ s_n = s_{n-1} + 2a n + (b - a) } \] where \( s_1 = a \cdot 1^2 + b \cdot 1 + c = a + b + c \). --- **Conclusion:** Every quadratic sequence (\( a n^2 + b n + c \)) has a recursion where the sequence of differences is arithmetic: \[ s_n = s_{n-1} + (2a n + b - a) \]

Certainly! Here are the concise final answers for both questions: --- ### **Question 2 Final Answer:** **Recursion relation for \( a_n = n^2 + 3n + 4 \):** \[ \boxed{ a_n = a_{n-1} + 2n + 2 } \] --- ### **Question 3 Final Answer:** **Recursion relation for the quadratic sequence \( s_n = a n^2 + b n + c \):** \[ \boxed{ s_n = s_{n-1} + 2a n + (b - a) } \]