work out. the \( n^{\text {th }} \) term for the 2 hadratic sequence \( 7,16,31,52,79 \)

To find the \( n^{\text{th}} \) term for the given quadratic sequence \( 7, 16, 31, 52, 79 \), we first need to identify the pattern in the sequence. The given sequence is a quadratic sequence, which means the difference between consecutive terms is not constant. To find the quadratic sequence, we need to determine the coefficients of the quadratic function that generates the sequence. Let's denote the quadratic function as \( f(n) = an^2 + bn + c \), where \( a \), \( b \), and \( c \) are constants to be determined. Given the first three terms of the sequence: 1. \( f(1) = 7 \) (when \( n = 1 \)) 2. \( f(2) = 16 \) (when \( n = 2 \)) 3. \( f(3) = 31 \) (when \( n = 3 \)) We can set up a system of equations to find the values of \( a \), \( b \), and \( c \) in the quadratic function. 1. \( a(1)^2 + b(1) + c = 7 \) 2. \( a(2)^2 + b(2) + c = 16 \) 3. \( a(3)^2 + b(3) + c = 31 \) Solving this system of equations will give us the values of \( a \), \( b \), and \( c \), which will allow us to find the \( n^{\text{th}} \) term of the sequence. Let's solve the system of equations to find the values of \( a \), \( b \), and \( c \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}a+b+c=7\\4a+2b+c=16\\9a+3b+c=31\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=7-b-c\\4a+2b+c=16\\9a+3b+c=31\end{array}\right.\) - step2: Substitute the value of \(a:\) \(\left\{ \begin{array}{l}4\left(7-b-c\right)+2b+c=16\\9\left(7-b-c\right)+3b+c=31\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}28-2b-3c=16\\63-6b-8c=31\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}b=\frac{12-3c}{2}\\63-6b-8c=31\end{array}\right.\) - step5: Substitute the value of \(b:\) \(63-6\times \frac{12-3c}{2}-8c=31\) - step6: Simplify: \(27+c=31\) - step7: Move the constant to the right side: \(c=31-27\) - step8: Subtract the numbers: \(c=4\) - step9: Substitute the value of \(c:\) \(b=\frac{12-3\times 4}{2}\) - step10: Simplify: \(b=0\) - step11: Substitute the values of \(b,c:\) \(a=7-0-4\) - step12: Simplify: \(a=3\) - step13: Calculate: \(\left\{ \begin{array}{l}a=3\\b=0\\c=4\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}a=3\\b=0\\c=4\end{array}\right.\) - step15: Rewrite: \(\left(a,b,c\right) = \left(3,0,4\right)\) The solution to the system of equations is \( a = 3 \), \( b = 0 \), and \( c = 4 \). Therefore, the quadratic function that generates the sequence is \( f(n) = 3n^2 + 4 \). To find the \( n^{\text{th}} \) term of the sequence, we substitute \( n \) into the quadratic function \( f(n) = 3n^2 + 4 \). So, the \( n^{\text{th}} \) term for the given quadratic sequence is \( 3n^2 + 4 \).