SERIES—CUBES AND OTHERS 35

then the original root number will appear (thus, 998 +o0o01
=999)-

Now the square number 81 can also be obtained from the
expression (8 X 10)+ 1, and this expression can be put into
terms of the root number (9) thus:

81=(9—1)(9+1)+1

Similarly 9801=(99—1)(99+1)+1
Substituting x2 for 81 or 9801, we obtain:
#2=(x—1)(x+1)+1

and we can immediately prove that the square of any num-
ber can be obtained by adding 1 to the multiple of the next
number below and the next above the original number, for
the equation x#?>=(x—1)(x+1)-+1 is the same as x?—1I
=(x—1)(x+1) and this is true for all values of x.

The cubes of ‘all-nine’ numbers are also of a pattern.

9°=729
99°=970299
999°=997002999

For each successive cube, we take the previous cube, add
another digit g at the front and at the end, and add a further
nought between the digits 7 and 2. The reason for this is not
immediately obvious, but a study of the following elementary
cube numbers uncovers the relationship.

23=8
3¥=27
43=064
53=125

These may be written differently:

28= 8= o+(3x2)+2
%= 27= 16+(3x3)+2
4°= 64= 50-+(3x4)+2
53=125=108-+(3 X5)+2
To begin with we may assume that each cube is partly
built up from three times its root number, plus 2. The original

 

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