Question

find the 5th term sequence defined as a(n)=(n+3)n-1

Answer 1

$a_{n} =(n+3)n-1 a_{1} =(1+3)*1-1=4-1=3 a_{2}=(2+3)*2-1=10-1=9 a_{3}=(3+3)*3-1=18-1=17 a_{4}=(4+3)*4-1=28-1=27 a_{5}=(5+3)*5-1=40-1=39$