Why are the perfect squares generatable by adding a succession of odd numbers?

0 is a perfect square. Add the first odd number to get 1. Add the next odd number to 1 and get 4. Add the next odd number to 4 and get 9.





Why does this work? What do odd numbers have to do with perfect squares?Why are the perfect squares generatable by adding a succession of odd numbers?
Two ways to see this:





Here is a formal proof:





The sum S of the first n odd numbers is





S = 1 + 3 + 5 + ......+(2n-5) + (2n-3) + (2n-1)





(for example, if n = 4, then S = 1 + 3 + 5 + 7, where 7 = 2(4) - 1)





Now we can rewrite this as





S = (2n-1) + (2n - 3) + (2n - 5 ) + .....+ 5 + 3 + 1





Add the two expressions together and you get





2S = [1+ 2n -1] + [ 3 + 2n - 3] + [ 5 + 2n - 5 ] + ....+ [2n -1 + 1]





2S = 2n + 2n + 2n + .....+ 2n (there are n terms)





S = n + n + n + ......n ( there are n terms)


= n*n= n^2.





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Here is an intuitive geometric argument.





Draw squares with dots:





*











**


**











***


***


***





for 1, 4, 9, etc.





now draw demarcations in the shape of inverted L's, to see each of these squares in ';layers';. For example, the last square, 9, can be thought of as an outer layer of 5 dots (across the top and right side) plus an inner 2x2 square which can be thought of as an outer layer of 3 dots, plus an inner single dot. 9 = 5 + 3 + 1.





and so on with the larger squares.Why are the perfect squares generatable by adding a succession of odd numbers?
Summations:





1+3+5+7+9...2n-1=(n)^2





So if n=4:





1+3+5+7=(4)^2=16





And to test this always works:





1+3+5+7...2n-1=n^2





Take the next term:





1+3+5+7...+(2n-1)+


(2n+1)=(n^2)+(2n+1)





And because:





(n^2)+(2n+1)=


(n+1)^2





You know that it is always a perfect square.





(This is a discrete mathematics problem)