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For example, if I had a perfect square of $16$, which is the fourth perfect square, I would add nine to get to the fifth perfect square, $25$. This is probably how it goes:$$1+3=4+5=9+7=16+9=25+11=36+13=49+15=64+17=81+19=100$$and it goes on forever. It looks like the amount being added each time gets added by two each time as they move on to the next and the next and the next and the next.

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First, think of the set of numbers to the left of each equal sign as a member of a sequence. $\{1+3=4, 4+5=9, 9+7=16, 16+9=25,...a_n\}$ There's a pattern on the left side:$$a_{n_L}=n^2+2n+1$$The right side is $a_{n_R}=(n+1)^2$. Expanding this results in $n^2+2n+1$. Does that look familiar?

Edit: I just realized this thread was a duplicate question. Oops...

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Two ways of proving that:
- by induction
- by drawing something like: http://reflectionsinthewhy.files.wordpress.com/2011/09/sum-of-consecutive-odd-numbers.jpg?w=584&h=902

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