SHARE I recently stumbled upon a brief blog post which demonstrated an interesting phenomenon.

The square of the sum of the series of consecutive numbers from 1 is equal to the sum of the cubes of the numbers.

The article gave specific examples:-

(1+2+3)2 = 13+23+33

and

(1+2+3+4)2 = 13+23+33+43

The above can, of course, be verified: (1+2+3)2 = 62 = 1+8+27; (1+2+3+4)2 = (1+2+3)2 + 43 = 36+64 = 100 = (1+2+3+4)2.

This extends to (1+2+3+4+5)2; 13+23+33+43+53 = (1+2+3+4)2 + 53 = 100 + 125 = 225 = (1+2+3+4+5)2.

A satisfying blog post, and thought provoking. However, it stopped short of providing a general proof for an arbitrary natural number n.

Well, here goes.

I start with the assertion that there exists a function, f(x), which is expressed as:-

f(x) = 1+2+3+4+5+6+7 …

Better expressed as:-

There must exist a function, g(x), such that

g(x) = (1+2+3+…)2

or

Finally, there exists a function, h(x), such that

h(x) = 13+23+33+…

or

(1+2+3+…)2 = 13+23+33+…

or

For a given arbitrary integer n, if g(x) = h(x) for x=n, then

(1+2+3+…+(n-2)+(n-1)+n)2 = 13+23+33+…+(n-2)3+(n-1)3+n3

Since 13+23+33+…+(n-2)3+(n-1)3 must be equivalent to (1+2+3+…+(n-2)+(n-1))2,

f(n) = (1+2+3+…+(n-2)+(n-1))2 + n3

must also be true.

If that is so, then it can also be said that

f(n) = (1+2+3+…+(n-2))2 + (n-1)3 + n3

is true, and

f(n) = (1+2+3+…+(n-3))2 + (n-2)3 + (n-1)3 + n3

is true, and so on.

This process can be repeated until we reach a number which we have already proven – in this case, 1+2+3+4+5, which we can assert as true:-

(1+2+3+4+5)2 = 13+23+33+43+53

To be certain, though, we can take this to the extreme case, f(1):-

If g(2) is equal to h(1) + 23

and h(1) = 13

then g(1) must be 12,

and since

12 = 13

then g(2)=h(2) must be true, which means g(3)=h(3) must be true … up to g(n)=h(n).

Therefore, for any arbitrary natural positive integer n,

I hope this proof meets with your approval. I submit it to the house.