Consider the quadratic number pattern: \( 2 ; 3 ; 6 ; 11 ; \ldots \) \( \begin{array}{ll}\text { (1) Determine the } 5 \text { th term. } & \text { (2) Determine the general } \\ \text { (3) Determine the } 50 \text { th term. } & \text { (4) Which term is equal to } \\ \text { Given the quadratic number pattern: } 3 ; 3 ; 4 ; 6 ; \ldots \text { Determine } \\ \text { (1) a formula for the } n \text {-th tern. } & \text { (2) } T_{\text {tix }}\end{array} \)

To determine the 5th term in the quadratic number pattern \(2; 3; 6; 11; \ldots\), we need to find the general formula for the nth term. Given the quadratic number pattern \(2; 3; 6; 11; \ldots\), we can observe that the differences between consecutive terms are increasing by 1 each time. This indicates that the pattern follows a quadratic relationship. Let's denote the nth term as \(T_n\). We can start by finding the differences between consecutive terms: - \(T_2 - T_1 = 3 - 2 = 1\) - \(T_3 - T_2 = 6 - 3 = 3\) - \(T_4 - T_3 = 11 - 6 = 5\) From the differences, we can see that the pattern of differences is increasing by 2 each time. This suggests that the quadratic relationship can be represented as \(T_n = an^2 + bn + c\), where \(a\), \(b\), and \(c\) are constants to be determined. Using the given terms, we can set up a system of equations to solve for \(a\), \(b\), and \(c\): 1. \(T_1 = 2 = a(1)^2 + b(1) + c\) 2. \(T_2 = 3 = a(2)^2 + b(2) + c\) 3. \(T_3 = 6 = a(3)^2 + b(3) + c\) Solving this system of equations will give us the values of \(a\), \(b\), and \(c\), which we can then use to find the general formula for the nth term. Let's proceed with solving 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: \(\left\{ \begin{array}{l}2=a+b+c\\3=4a+2b+c\\6=9a+3b+c\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=2-b-c\\3=4a+2b+c\\6=9a+3b+c\end{array}\right.\) - step2: Substitute the value of \(a:\) \(\left\{ \begin{array}{l}3=4\left(2-b-c\right)+2b+c\\6=9\left(2-b-c\right)+3b+c\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}3=8-2b-3c\\6=18-6b-8c\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}b=\frac{5-3c}{2}\\6=18-6b-8c\end{array}\right.\) - step5: Substitute the value of \(b:\) \(6=18-6\times \frac{5-3c}{2}-8c\) - step6: Simplify: \(6=3+c\) - step7: Swap the sides: \(3+c=6\) - step8: Move the constant to the right side: \(c=6-3\) - step9: Subtract the numbers: \(c=3\) - step10: Substitute the value of \(c:\) \(b=\frac{5-3\times 3}{2}\) - step11: Simplify: \(b=-2\) - step12: Substitute the values of \(b,c:\) \(a=2-\left(-2\right)-3\) - step13: Simplify: \(a=1\) - step14: Calculate: \(\left\{ \begin{array}{l}a=1\\b=-2\\c=3\end{array}\right.\) - step15: Check the solution: \(\left\{ \begin{array}{l}a=1\\b=-2\\c=3\end{array}\right.\) - step16: Rewrite: \(\left(a,b,c\right) = \left(1,-2,3\right)\) The solution to the system of equations is \(a = 1\), \(b = -2\), and \(c = 3\). Therefore, the general formula for the nth term in the quadratic number pattern is \(T_n = n^2 - 2n + 3\). Now, we can use this formula to find the 5th term: \(T_5 = 5^2 - 2(5) + 3 = 25 - 10 + 3 = 18\). So, the 5th term in the quadratic number pattern is 18. Next, we need to determine the 50th term using the same formula: \(T_{50} = 50^2 - 2(50) + 3 = 2500 - 100 + 3 = 2403\). Therefore, the 50th term in the quadratic number pattern is 2403. Finally, we need to find the term that is equal to 2403 using the same formula: \(2403 = n^2 - 2n + 3\). Solving for \(n\), we get: \(n^2 - 2n + 3 = 2403\) \(n^2 - 2n - 2400 = 0\) Solving this quadratic equation, we find that \(n = 50\). Therefore, the term that is equal to 2403 is the 50th term in the quadratic number pattern.