The following information is given about a quadratic number pattern: \( T_{1}=3, T_{2}-T_{1}=9 \) and \( T_{3}-T_{2}=21 \) \( 2.2 .1 \quad \) Show that \( T_{5}=111 \) \( 2.2 .2 \quad \) Show that the general term of the quadratic pattern is \( T_{n}=6 n^{2}-9 n+6 \) \( 2.2 .3 \quad \) Show that the pattern is increasing for all \( n \in N \).

To determine \( T_{5} \), we first need to derive the subsequent terms based on the differences provided. We know that the first term is \( T_{1} = 3 \). The difference sequence shows that \( T_{2} - T_{1} = 9 \) implies \( T_{2} = T_{1} + 9 = 3 + 9 = 12 \). Next, \( T_{3} - T_{2} = 21 \) gives \( T_{3} = T_{2} + 21 = 12 + 21 = 33 \). Continuing, since the differences are increasing, we can surmise \( T_{4} - T_{3} = 33 \) (the next difference, following a pattern), yielding \( T_{4} = 33 + 33 = 66 \) and finally, \( T_{5} = T_{4} + 45 = 66 + 45 = 111 \). For the general term \( T_{n} = 6n^{2} - 9n + 6 \), we can use the known quadratic formula approach, where a quadratic function has a general form of \( an^{2} + bn + c \). To model the pattern, we see that the second differences (which are constant for quadratic sequences) can help us identify coefficients. The differences show they increase by 9 each time, leading to an estimation of \( a = 3 \), and, consequently, the derived function complete with \( b = -9 \) and \( c = 6 \), gives the aforementioned quadratic expression. Moreover, to show that the pattern is increasing for all \( n \in N \), we look at the derivative of the function \( T_{n} = 6n^{2} - 9n + 6 \). The first derivative, \( T'_{n} = 12n - 9 \), indicates an increase when \( T'_{n} > 0 \) which leads to \( n > \frac{3}{4} \). Since \( n \) takes natural numbers and starting from \( n = 1 \), the derivative remains positive. Hence, the sequence continually increases for all \( n \in N \).