MATHEMATICAL RECREATIONS 189

is a number consisting of r — 1 ones, or

r—3
[Z k-r"’“”—l—(r—])](r—l) =111---111 (r— 1 ones, radix r)
k=1

r—3
To prove this theorem let N = [ 2 kerF 24 (r—1) ](r— 1)
k=1

r—3
The terms of > k-7"~*=2 are r — 3 in number.
k=1
Hence the kth term from the beginning is the
(r — 3) — k + 1th, or » — k — 2th, term from the end.
Of course the sum is unalterd by reversing the order of the
terms, that is, by replacing 2 by » — & — 2.

Thus N=[T§_‘,3(r—k—2)r’“—|—(r—1)](1'—1)

G ECoD PG

r—3
Hence N=14+ (@ —-2)r+ @ —1) > (r — k — 2)r*
k=1
This statement is a special case of the statement
N=14+1r+1r24 -« + 14 (r — (I + 1))
r—3
etk W~ =0
k=1
where 1 = ] = » — 3, the case when / = 1.
We will prove that all the statements obtaind from this form
by giving / the integral values from 1 to r — 3 inclusiv are tru.
We have seen that the statement obtaind by putting / = 1

is tru.
Suppose that any one of these statements except the last is

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