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Question 265111: Why does a perfect square result from adding two successive rows of even and odd integers? Example: 1+2 plus 1+2+3 = 9
1+2+3 plus 1+2+3+4 = 16
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let n = last number in first row
If we generalize the problem, we get
1+2+3+...+n
1+2+3+...+n+(n+1)
Add the two rows to get
(1+2+3+...+n)+(1+2+3+...+n+(n+1))
Now simplify:
(1+1)+(2+2)+(3+3)+...+(n+n)+n+1
2(1)+2(2)+2(3)+...+2(n)+n+1
2(1+2+3+...+n)+n+1
Now use the formula 1+2+3+...+n = n(n+1)/2
2(n(n+1)/2)+n+1
n(n+1)+n+1
n^2+n+n+1
n^2+2n+1
(n+1)^2
So this shows that the sum of 1+2+3+...+n and 1+2+3+...+n+(n+1) is a perfect square and it shows us that the perfect square is (n+1)^2
In the case that n = 4, we get
(1+2+3+4)+(1+2+3+4+5) = 10+15 = 25
Take note that 25 = 5^2 = (4+1)^2 which confirms our answer.
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