Induction Sum of all odd Numbers

Question

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?

Answer 1

Induction step:

Assume $\sum_{k=1}^n$ $(2k - 1)$ = $n^2$. Then $\sum_{k=1}^{n+1}$ $(2k - 1)$ = $\sum_{k=1}^n$ $(2k - 1)$ + [$(2(n + 1) - 1)$] = $n^2 + [2(n + 1) - 1]$ = $n^2 + 2n + 1$ = $(n + 1)^2$.

Answer 2

Hint:

$1$ $ +3+5+\cdots+2k-1=x$

$2k-1+2k-3++\cdots+3+1=x$

$2k+2k+\ldots+2k=2x$