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$1 + 3 = 4$ (or $2$ squared)
$1+3+5 = 9$ (or $3$ squared)
$1+3+5+7 = 16$ (or $4$ squared)
$1+3+5+7+9 = 25$ (or $5$ squared)
$1+3+5+7+9+11 = 36$ (or $6$ squared)

you can go on like this as far as you want, and as long as you continue to add odd numbers in order like that, your answer is always going to be a perfect square.

But how to prove it?

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    $\begingroup$ Hint: $(n+1)^2 = n^2 + (2n + 1)$, ie, squaring the next number adds the corresponding odd number. $\endgroup$ – Tobias Kildetoft Jan 15 '14 at 9:44
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\begin{align*} \sum_{k=0}^{n}[2k+1] &= \sum_{k=0}^{n}[2k] + \sum_{k=0}^{n}1=\sum_{k=0}^{n}[2k] + (n+1)\\ &=2\sum_{k=0}^{n}[k] + (n+1)= 2\frac{n(n+1)}{2}+(n+1) = (n+1)^2 \end{align*}

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Consider a square. It can be divided as the following:

square divided into odd numbers

Each red segment contains odd number of circles.

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    $\begingroup$ Nice geometric proof! $\endgroup$ – A.P. Jan 15 '14 at 9:53
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    $\begingroup$ Outrageous..... $\endgroup$ – Mark C Oct 4 '17 at 9:19
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Because $$\begin{align}1+3+5+\cdots+(2n-1)&=\sum_{k=1}^{n}(2k-1)\\&=2\sum_{k=1}^{n}k-\sum_{k=1}^{n}1\\&=2\cdot\frac{n(n+1)}{2}-n\\&=n(n+1)-n\\&=n^2.\end{align}$$

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You can prove it by induction . Statement : $ 1+3+5+...+(2n-1)=n^2 $

Base case : For n =1 , the LHS of statement is 1 and the RHS of the statement is 1 . So the statement is true for n=1 . Induction step : Let the statement is true .

$1+3+5+...... + (2n-1)=n^2$

So $ 1+3+5+...+(2n-1)+(2n+1) = n^2+2n+1 = (n+1)^2 $

So the statement is true for all natural values of n .

$1+3+5+...+(2n-1)=n^2 $

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