If [tex]n=1[/tex], then [tex]1=1^2[/tex] is true.
Suppose the equation holds for [tex]n=k[/tex], so that
[tex]1+3+5+\cdots+(2k-1)=k^2[/tex]
Then if [tex]n=k+1[/tex], we have
[tex]1+3+5+\cdots+(2k-1)+(2(k+1)-1)=k^2+(2k+1)=(k+1)^2[/tex]
as required, so the equation holds for all [tex]n\in\mathbb N[/tex].
