I come recently across this nice result. Displayed that way, it looks like a poem. Too nice to be true, so simple in its beauty!
Proof:
A simple induction is sufficient to prove it. We consider the following induction hypothesis.
P(n): (1+2+â¦+n)2 = 13 + 23+â¦+ n3
We verify that P(1) is true:
12 = 1 = 13
So P(1) is true.
Now, suppose P(n) is true. Then
Now, suppose P(n) is true. Then
[1+2+â¦+n+(n+1)]2 = (1+2+...+n)2 + 2(1+2+...+n)(n+1)+(n+1)2 = 13+23+...+n3 + (n+1)[2(1+2+...+n)+(n+1)] (as P(n) is supposed true) = 13+23+...+n3 + (n+1)[2 n(n+1)/2 + (n+1)] (using the sum of the n first integer, result which also can be proved by induction) = 13+23+...+n3 + (n+1)[n(n+1) + (n+1)] = 13+23+...+n3 + (n+1)3
Proving that P(n+1) is true whenever P(n) is true. So for all positive integer n,
(1+2+â¦+n)2 = 13 + 23+â¦+ n3
A last word...
Next time I teach induction, I will definitely use that result.... Math is also poetry: