(a) Notice that:
[tex]\begin{gathered} -2=1^2-3, \\ 1=2^2-3, \\ 6=3^2-3, \\ 13=4^2-3, \\ 22=5^2-3. \end{gathered}[/tex]
Therefore a closed-form formula for an is:
[tex]a_n=n^2-3.[/tex]
(b) Notice that:
[tex]\begin{gathered} 1=2\cdot2-1+(-2), \\ 6=2\cdot3-1+1, \\ 13=2\cdot4-1+6, \\ 22=2\cdot5-1+13. \end{gathered}[/tex]
Therefore a recursive formula for an is:
[tex]\begin{gathered} a_n=2n-1+a_{n-1}, \\ a_1=-2. \end{gathered}[/tex]
Answer:
(a)
[tex]a_n=n^2-3._{}[/tex]
(b)
[tex]\begin{gathered} a_n=2n-1+a_{n-1}, \\ a_1=-2. \end{gathered}[/tex]
