<s>[INST] Induction Sum of all odd Numbers 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?
 [/INST] Induction step:
Assume $\sum_{k=1}^n$ $(2k - 1)$ = $n^2$. Then $\sum_{k=1}^{n+1}$ $(2k - 1)$ = $\sum_{k=1}^n$ $(2k - 1)$ + [$(2(n + 1) - 1)$] = $n^2 + [2(n + 1) - 1]$ = $n^2 + 2n + 1$ = $(n + 1)^2$.
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