SERIES—CUBES AND OTHERS 29

nth triangular number. Since we have shown that each

triangular number is of the form g(n +1), it follows that each

hexagonal number is of the form 62—72(72 +1)+1=3n24+3nt1,

and this expression is equal to the difference between two
consecutive cubes.

(n4-1)3=n3+3gn2tgn-t1
and (n+1)3—n3=gn24-gn-+1
and this proves that the summation of a series of hexagonal
numbers will give the relative cube number.

Since all square numbers may be built up from series of
odd numbers from 1 upwards, it follows that the addition of
any cubes, whose series together use up all consecutive odd
numbers rising from 1 upwards, will give a square number;
and the root of that square will be equivalent to the sum of
the roots of the cube numbers.

Thus, 14+243=06

and 134234-33=62=(1+2-3)2
(14+8-+27)=36

This is because:

I = J=—13
3%5 2822
TS5 53 =27=33

I1+3+5+7+9+11=36=62

This relationship eliminates the separate calculation of
each individual cube number in any larger calculation
which requires the summation of a series of consecutive cube
numbers rising from 12 For example, the series 13423493
t+ ... +25%sumsto (1+2+34 . ..25)%,which is the square
of the sum of an arithmetical progression, and may therefore
be shown as

[MT__(% X 26

2 N

 

2
) =325%=105,625

L . AT S ANAAHIRO B

BRI RER R eROs

&
3