SERIES—SHAPES AND SQUARES 21

Thus the sum of the grd and 2nd triangular numbers
(643) =9=32 and the sum of the 5th and 4th numbers
(15+10) =25 =52 This relationship can be shown in another
way:

1%351 =(1)

2?=1+43  =(1)+(142) .

3*=3+6  =(1+42)+ (+2+3)

4*=6+10 =(1+2+3)+(1+2+43+4)
5}=10+15 =(1+2+3+4)+(1+2+3+4+5)

and this shows clearly the nature of the two triangular num-
bers of which each square number is composed. These
expressions can be simplified still further. Thus the expression
for 42, which is (14+2+4-3) 4+ (1+2+3+4) can be reduced to
2(1+2+3) +4. Substituting x for 4, we obtain

x?=2(Sum of 1 to x—1) +x

=2<§>(x—1)—|—x

=(x)(x—1) +x
=x2—x+x=x2
thus proving the proposition.
Square numbers can also be built up by the addition of
Arithmetical Progressions containing all the consecutive odd
numbers from 1 upwards, as follows:

143 = 4 =22
I1+3+5 =9 =32
1+3+5+7 =16 =42
1+3+5+7+9 =25 =52

It is also observable that the number of terms in each pro-
gression is the same as the number whose square results (thus
the square of 6 has 6 terms in its progression). Furthermore,
the last term in any progression is equal to twice the number
to be squared, minus 1. (The last term for 62=2 X6 —1=11.)
Thus all squares may be calculated by the addition of the
relevant series of odd numbers, and there are two ways of
checking where each series ends.

 

st e cosRRRs a3

i

AR N

s RS

GRS St

5
e
=

:-l

]
i
B

o

R OBHKURR R BRI
S A AT RS