Answer :
[tex] a_{n} =(n+3)n-1
a_{1} =(1+3)*1-1=4-1=3
a_{2}=(2+3)*2-1=10-1=9
a_{3}=(3+3)*3-1=18-1=17
a_{4}=(4+3)*4-1=28-1=27
a_{5}=(5+3)*5-1=40-1=39[/tex]
a_{1} =(1+3)*1-1=4-1=3
a_{2}=(2+3)*2-1=10-1=9
a_{3}=(3+3)*3-1=18-1=17
a_{4}=(4+3)*4-1=28-1=27
a_{5}=(5+3)*5-1=40-1=39[/tex]
