(i) ( a_n = 3n + 5 )

[ \begin{align*} a_1 &= 3(1) + 5 = 8 \ a_2 &= 3(2) + 5 = 11 \ a_3 &= 3(3) + 5 = 14 \end{align*} ]

Answer: ( 8, 11, 14 )

(ii) ( a_{n+1} = 4a_n - 7 ), ( a_1 = 3 )

[ \begin{align*} a_2 &= 4(3) - 7 = 5 \ a_3 &= 4(5) - 7 = 13 \end{align*} ]

Answer: ( 3, 5, 13 )

(iii) ( a_n = (n-3)(n+1) )

[ \begin{align*} a_1 &= (1-3)(1+1) = (-2)(2) = -4 \ a_2 &= (2-3)(2+1) = (-1)(3) = -3 \ a_3 &= (3-3)(3+1) = (0)(4) = 0 \end{align*} ]

Answer: ( -4, -3, 0 )

(iv) ( a_1 = -1 ), ( a_{n+1} = \dfrac{3}{a_n + 2} )

[ \begin{align*} a_2 &= \dfrac{3}{-1 + 2} = \dfrac{3}{1} = 3 \ a_3 &= \dfrac{3}{3 + 2} = \dfrac{3}{5} \end{align*} ]

Answer: ( -1, 3, \dfrac{3}{5} )

(v) ( a_n = 8 - \dfrac{20}{3 + n} )

[ \begin{align*} a_1 &= 8 - \dfrac{20}{3+1} = 8 - 5 = 3 \ a_2 &= 8 - \dfrac{20}{3+2} = 8 - 4 = 4 \ a_3 &= 8 - \dfrac{20}{3+3} = 8 - \dfrac{20}{6} = 8 - \dfrac{10}{3} = \dfrac{14}{3} \end{align*} ]

Answer: ( 3, 4, \dfrac{14}{3} )

(vi) ( a_1 = 1 ), ( a_{n+1} = (3a_n + 2)^2 )

[ \begin{align*} a_2 &= (3\cdot1 + 2)^2 = 5^2 = 25 \ a_3 &= (3\cdot25 + 2)^2 = 77^2 = 5929 \end{align*} ]

Answer: ( 1, 25, 5929 )

(vii) ( a_n = (-2n)^2 )

[ \begin{align*} a_1 &= (-2\cdot1)^2 = 4 \ a_2 &= (-2\cdot2)^2 = 16 \ a_3 &= (-2\cdot3)^2 = 36 \end{align*} ]

Answer: ( 4, 16, 36 )

(viii) ( a_n = (-1)^n 7n^2 )

[ \begin{align*} a_1 &= (-1)^1 \cdot 7 \cdot 1^2 = -7 \ a_2 &= (-1)^2 \cdot 7 \cdot 4 = 28 \ a_3 &= (-1)^3 \cdot 7 \cdot 9 = -63 \end{align*} ]

Answer: ( -7, 28, -63 )