The nth square number is given by the formula: Sn= n
×n= n2. Looking at the diagonals of the square reveals
the sum:
1 + 2 +…+ (n– 1) + n+ (n– 1) +…+ 2 + 1 = n2
Also hidden in a square is the sum of the first nodd
numbers. We thus have:
The sum of the first nodd numbers is n2.
Adding one to each of these summands gives the sum
of the first neven numbers. Thus we have:
The sum of the first neven numbers is n2+ n.
There is a nice interplay between the triangular and
square numbers. For example, the sum of any two con-
secutive triangular numbers is always a square number:
1 + 3 equals 4, 3 + 6 equals 9, and 6 + 10 equals 16,
for instance. In general:
Tn–1 + Tn= Sn
The center diagram in the figure at the bottom of page
194 explains why this is the case.
Similarly, one can arrange eight copies of the one
triangle to form a square with its center removed to
prove that 8Tnis always one less than a square number.
In the same way, one can also establish that the follow-
ing combination of three consecutive triangular num-
bers is always square: Tn–1 + 6Tn+ Tn+1.
The numbers 1 = T1= S1, 36 = T8= S6and 1225 =
T49 = S35 are both square and triangular, as are 41,616
and 1,413,721. Mathematicians have proved, using
CONTINUED FRACTION
s, that there are infinitely num-
bers with this property.
The alternate triangular numbers 1, 6, 15, 28, …
are sometimes called the Bohlen numbers. The nth
Bohlen number xis divisible by nand is the unique
multiple of nsatisfying: 1 + 2 +…+ = x.
In 1796 K
ARL
F
RIEDRICH
G
AUSS
proved that every
natural number is the sum of at most three triangular
numbers, and J
OSEPH
-L
OUIS
L
AGRANGE
in 1770 proved
that every natural number is the sum of no more than
four square numbers.
One can extend the scope of figurative numbers to
three dimensions to produce the cube numbers, tetrahe-
dral numbers, and the like. The nth cube number is
given by the formula n3. It is a three-dimensional cubi-
cal array of dots, with each layer being a square array
of dots. (Thus nlayers of n2dots gives a total of n×n2
= n3elements.) The nth tetrahedral number is produced
by stacking together the first ntriangular numbers:
T1+ T2+…+ Tn. This gives the sequence:
1, 4 = 1 + 3, 10 = 1 + 3 + 6, 20 = 1 + 3 + 6 + 10,…
One can prove that the nth tetrahedral number is given
by the formula n(n+ 1)(n+2).
Both the triangular numbers and the tetrahedral
numbers appear as diagonals in P
ASCAL
’
S TRIANGLE
. The
successive stacking of tetrahedral numbers produces
hypertetrahedral numbers: 1, 5, 15, 35, … These corre-
spond to four-dimensional geometric arrangements of
dots. They also appear as a diagonal in Pascal’s triangle.
See also
NESTED MULTIPLICATION
;
PERFECT NUMBER
;
SQUARE
;
TRIANGLE
.
finger multiplication Having memorized the 2-, 3-,
4-, and 5-times tables, there is a popular finger met-
hod for computing all values of the 6- through 10-
times tables. It is based on the following rule for
encoding numbers:
A closed fist represents 5 and any finger raised
on that hand adds 1 to that value.
Thus a hand with one finger raised, for example, repre-
sents 6. A hand with three fingers raised represents 8.
To multiply two numbers between 5 and 10, one then
follows these steps:
1. Encode the two numbers, one on each hand.
2. Count 10 for each finger raised.
3. Count the number of unraised fingers on each hand
and multiply together the two counts.
4. Add the results of steps two and three. This is the
desired product.
For example, “6 times 8” is represented as one raised
finger on the left hand, three on the right hand. There
are four raised fingers in all, yielding the number 40 for
step 2. The left hand has four lowered fingers, and the
right has two fingers lowered. We compute 4 ×2 = 8.
Thus the desired product is 40 + 8 = 48. Similarly, “8
1
–
6
x
–
n
finger multiplication 195