What are the first four solutions to the sequence: f (1) = 4 and f (n) = 2*f (n-1) + 3

F (1) = 4 is given

f (2) = 2*f (2-1) + 3

= 2*f (1) + 3

= 2*4+3

= 8+3

= 11

f (3) = 2*f (3-1) + 3

= 2*f (2) + 3

= 2*11+3

= 22+3

= 25

f (4) = 2*f (4-1) + 3

= 2*f (3) + 3

= 2*25+3

= 50+3

= 53

f (5) = 2*f (5-1) + 3

= 2*f (4) + 3

= 2*53+3

= 106+3

= 109

f (1) = 4

f (2) = 11

f (3) = 25

f (4) = 53

f (5) = 109