Show that

$\sum\_{k=1}^{n}(2k-1)=n^2$

Beginning: n=1

$\sum\_{k=1}^{1}(2k-1)=(2\*1-1)=1=1^2$

Let $\sum\_{k=1}^{n}(2k-1)=n^2$ be true, then for n=p+1

$\sum\_{k=1}^{p+1}(2k-1)=(p+1)^2$ has to be true too.

$\sum\_{k=1}^{p+1}(2k-1)=1+3+...+(2(p+1)-3)+(2(p+1)-1)$

$=1+3+...+(2p-1)+(2p+1)$

$=1+3+...+(2p-3)+(2p-1)+(2p+1)$

(Using our assumption that $1+3+...+(2p-3)+(2p-1)=p^2$)

$=p^2+2p+1$

$=(p+1)^2$ which is exactly what we wanted to show.

Is this ok?