TITLE: Square root and sum of odd numbers
QUESTION [5 upvotes]: I have found this on internet as a very interesting mathematical curiosity...

If we have $x\in \mathbb{N}$, they say that if $x=\sum^{K}_{n=0}{(2n+1)}$, then $\sqrt{x}\in \mathbb{N}$ and $\sqrt{x}=K+1$.

Is that always true? Why?
This is my first question here! Thanks in advance and sorry for any mistakes.

REPLY [2 votes]: That is my favorite fact!
Consider this.
It's true that $1=1^2$
And it is true that $1+3=2^2$ and that $1+3+5=3^2$
Suppose it were true that $1+3+5+.....+(2n-1)=n^2$.
Then $1+3+5+.....+(2n-1)+(2n+1)=$
$n^2+(2n+1)= $
$n+2n+1=(n+1)^2$.
So if it's true for some it is true for the next.
Also.
Consider:
$N= 1+2+......+n $
Then $2N=N+N=$
$1+2+......+n+$
$n+......+2+1=$
$(n+1)+(n+1)+..... +(n+1) =$
$n (n+1) $
So $N=n (n+1)/2$
So $\sum (2k-1)=2\sum k -\sum 1$
$=2*\frac {n (n+1)}2-n$
$=n (n+1) - n=n^2+n-n=n^2$.
I love it.