Answer:
a).
[tex]{ \bf{first \: term}} \\ let \: n = 1 \\ n_{1} = {1}^{2} + 3(1) + 1 \\ n_{1} = 5 \\ \\ { \bf{second \: term}} \\ let \: n = 2 \\ n_{2} = {2}^{2} + 3(2) + 1 \\ n_{2} = 11 \\ \\ { \bf{third \: term}} \\ n_{3} = {3}^{2} + 3(3) + 1 \\ n_{3} = 19 \\ \\ { \bf{fourth \: term}} \\ n_{4} = {4}^{2} + 3(4) + 1 \\ n_{4} = 29[/tex]
b).
common difference, d:
[tex]d = n_{4} - n_{3} \\ d = 29 - 19 \\ d = 10\\ but \: increases \: by \: 2 \\ d = 10 + 2 = 12\\ therefore : \\ n_{5} = n_{4} + d \\ n_{5} = 29 + 12 \\ n_{5} = 41[/tex]
c).
[tex]n_{5} = {5}^{2} + 3(5) + 1 \\ n_{5} = 41[/tex]
yes it is the same
