SCALES OF NOTATION 31

Hence

k l k+1
o= 2@l o) + gl ()
=0 m=0 m=0
which is the £ 4 1th in the series of statements.

Therfor by mathematical induction the statements are
all tru.

If we do not wish to use the device that 7o = 1, the con-
clusion of this theorem may be put in the form

k
ifk=1,a6=a+ g1l
m=1

k-1 l k
e {almrm)} L alie)
=1 m=1

m=1
=1 m=1

k—1 1
The expression ) { a; IT (rm)} has no sense if & = 1.

Again, the conclusion is a special case, the case when j = 1,
of the statement

i+k—2 Jk j+k—1
qi—1 = Z {al 1I (rm)} + Qj+r—1 II (T’m), provided ri—1 = 1;
l=7—1 m=j—1 m=j—1
or, if we do not wish to use the device that ;_; = 1,
j+k—1
if k=1, ¢j1 = aj1 + giyr1 11 (7m)
m=j
7+k—2 l 7+k—1
ifk>1, =60+ % {azH (fm)} + g I (),
=3 m=j m=j

which may be proved in the same way.
The conclusion of our theorem for the first three values of
k gives the following results:

putting £ = 1, we have ¢ = ¢y = a0 + g1
putting & = 2, a = ao.+ a1 + qnire
putting & = 3, a = ag + a1 + aerirs + qarirers