Example 1: Given quadratic sequence is defined as \( T_{k}=\sum_{n=0}^{k-1}(6 n-2) \) 1. Show that the first 3 terms of the quadratic sequence are given by: \( -2 ; 2 ; 12 ; \ldots \ldots \)

To demonstrate that the first three terms of the quadratic sequence defined by \( T_{k} = \sum_{n=0}^{k-1} (6n - 2) \) are \(-2\), \(2\), and \(12\), we'll calculate each term individually by evaluating the sum for \( k = 1 \), \( k = 2 \), and \( k = 3 \). ### Calculating \( T_1 \) For \( k = 1 \): \[ T_1 = \sum_{n=0}^{0} (6n - 2) \] Since the upper limit is \( n = 0 \), the sum includes only one term (\( n = 0 \)): \[ T_1 = 6(0) - 2 = 0 - 2 = -2 \] ### Calculating \( T_2 \) For \( k = 2 \): \[ T_2 = \sum_{n=0}^{1} (6n - 2) \] This sum includes two terms (\( n = 0 \) and \( n = 1 \)): \[ \begin{align*} T_2 &= (6(0) - 2) + (6(1) - 2) \\ &= -2 + (6 - 2) \\ &= -2 + 4 \\ &= 2 \end{align*} \] ### Calculating \( T_3 \) For \( k = 3 \): \[ T_3 = \sum_{n=0}^{2} (6n - 2) \] This sum includes three terms (\( n = 0 \), \( n = 1 \), and \( n = 2 \)): \[ \begin{align*} T_3 &= (6(0) - 2) + (6(1) - 2) + (6(2) - 2) \\ &= -2 + 4 + (12 - 2) \\ &= -2 + 4 + 10 \\ &= 12 \end{align*} \] ### Summary of the First Three Terms \[ T_1 = -2 \\ T_2 = 2 \\ T_3 = 12 \\ \] Thus, the first three terms of the quadratic sequence are \(-2\), \(2\), and \(12\).