Question 265111
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.