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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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marked as duplicate by Bookend, user88595, Chris Janjigian, Michael Albanese, qwr Jun 6 '14 at 1:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Hint: $(n+1)^2 = n^2 + (2n + 1)$, ie, squaring the next number adds the corresponding odd number. – Tobias Kildetoft Jan 15 '14 at 9:44
up vote 1 down vote accepted

\begin{align*} \sum_{k=0}^{n}[2k+1] &= \sum_{k=0}^{n}[2k] + \sum_{n=0}^{k}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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Nice geometric proof! – A.P. Jan 15 '14 at 9:53

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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