Question

I need help with this problem and please show step by step

Answer 1

a).

`{ \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`

b).

common difference, d:

`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`

c).

`n_(5) = {5}^(2) + 3(5) + 1 \\ n_(5) = 41`

yes it is the same