Appendix 3 Section 2
The sum of consecutive cubes
When the same number is repeated as a factor three times -- as 4 × 4 × 4 -- we call the product the 3rd power of that base; that product is commonly called a cube. (This is analogous to the volume of the solid figure called a cube.)
Here is number 4:
Upon repeatedly adding it four times --
-- we have the 2nd power, or the square, of 4.
Upon repeatedly adding that power four times --
-- we have the 3rd power, or the cube, of 4.
It will be convenient for the moment to express the cube of a number with the exponent 3.
We come now to one of the most remarkable facts in the structure of the natural numbers:
The sum of n consecutive cubes is equal to the square
13 + 23 + 33 + . . . + n3 = (1 + 2 + 3 + . . . + n)2.
To see that, we will begin here:
The difference between the squares of two consecutive triangular numbers
Triangles: 1 3 6 10 15 21 28
The base of each cube is the difference of the two triangles.
Look -- here is the cube of 4:
From it, let us separate this rectangular array --
-- and reposition it here:
Then we have the square of side 10 --
-- minus the square of side 6.
The difference between the squares of those two consecutive triangles, 10 and 6, is equal to a cube, 43.
The sum of those four cubes is equal to the square of the fourth triangle.
Alternatively, since every square number is the sum of consecutive odd numbers, so is the square of a triangular number.
Therefore, the difference of those squares -- each cube -- will be a sum of consecutive odd numbers, although not starting with 1.
Again, the sum of those four cubes is equal to the square of the fourth triangle.
We proved that by mathematical induction
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