we have the sequence
5+9+13+...
we have an arithmetic sequence
a1=5
a2=9
a3=13
a2-a1=9-5=4
a3-a2=13-9=4
the common difference is d=4
the general expression is equal to
[tex]a_n=a_1+d\cdot(n-1)[/tex]
we have
a1=5
d=4
substitute
[tex]\begin{gathered} a_n=5+4\cdot(n-1) \\ a_n=4n+1 \end{gathered}[/tex]
therefore
the notation is equal to
see the attached figure
please wait a minute to fill the image
