Let [tex]S(n)[/tex] be the statement that
[tex]1+3+5+\cdots+(2n-1)=\displaystyle\sum_{k=1}^n(2k-1)=n^2[/tex]
Check to see that [tex]S(n)[/tex] holds for [tex]n=1[/tex]:
[tex]\displaystyle\sum_{k=1}^1(2k-1)=1=1^2\implies S(1)\text{ is true}[/tex]
Now suppose that [tex]S(N)[/tex] is true, and use that assumption to show that [tex]S(N+1)[/tex] must also be true. When [tex]n=N+1[/tex], you have
[tex]\displaystyle\sum_{k=1}^{N+1}(2k-1)=\sum_{k=1}^N(2k-1)+2(N+1)-1[/tex]
[tex]=N^2+2(N+1)-1[/tex]
[tex]=N^2+2N+1[/tex]
[tex]=(N+1)^2[/tex]
and so [tex]S(N+1)[/tex] is also true, thus proving the statement is true for all [tex]n[/tex].
