American Journal of Applied Science and Technology
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VOLUME
Vol.05 Issue01 2025
PAGE NO.
42-51
10.37547/ajast/Volume05Issue02-12
Use of AL-KARAJI method in calculating sum
Azizbek Kaxxorov
Tashkent State Technical University named after Islam Karimov, Uzbekistan
Akmal Shernayev
Tashkent State Technical University named after Islam Karimov, Uzbekistan
Received:
22 December 2024;
Accepted:
24 January 2025;
Published:
26 February 2025
Abstract:
This article provides several ways to find the sum of some number series. Al-Karaji's contribution to the
science of mathematics lies in the method that he used to calculate the sums, and finding of the sum of a series of
cubes of natural numbers is proved in this way. To explain al-Karaji's method more widely, other sums were
calculated in the same way, and a general result was obtained.
Keywords:
Number series, Newton binomial, binomial coefficient, arithmetic progression, short multiplication
formula, rectangle, square.
Introduction:
The development of Arab mathematics
began in the 7th century AD, just in the era of the
emergence of the religion of Islam. It grew out of the
many challenges posed by trade, architecture,
astronomy, geography, optics, and deeply combined
the desire to solve these practical problems and
intense theoretical work. Arab mathematicians
achieved significant achievements and made a number
of undeniable discoveries in the development of
algebraic calculus, both abstract and practical, the
formation of the theory of equations, algorithmic
methods at the junction of algebra and arithmetic. In
the development of Arab mathematics, two stages can
be distinguished: first of all, the assimilation in the 7th
and 8th centuries of the Greek and Eastern heritage.
Baghdad was the first major scientific center during the
reign of al-Mansur (754-775) and Harun al-Rashid (786-
809). There were a large number of libraries, and many
copies of scientific works were made. The works of
ancient Greece (Euclid, Archimedes, Apollonius, Heron,
Ptolemy, Diophantus) were translated, and works from
India, Persia and Mesopotamia were also studied. But
by the 9th century, a real Arab mathematical culture of
its own had formed, and new work went beyond the
limits defined by the Hellenic mathematical heritage.
The first famous scientist of the Baghdad school was
Muhammad al-Khwarizmi, whose activity took place in
the first half of the 9th century. He was part of a group
of mathematicians and astronomers who worked in the
House of Wisdom, a kind of academy founded in
Baghdad during the reign of al-Mammun (813-833).
Five works by al-Khorezmi have survived, partially
revised, of which two treatises on arithmetic and
algebra had a decisive impact on the further
development of mathematics. His treatise on
arithmetic is known only in the Latin version of the 13th
century, which, no doubt, is not an accurate
translation. It could be titled "A Book on Addition and
Subtraction Based on Indian Calculus." This is, in any
case, the first book that sets out the decimal number
system and the operations performed in this system,
including multiplication and division. In particular, a
small circle was used there, which served as a zero. Al-
Khorezmi explained how to pronounce numbers using
the concepts of one, ten, hundreds, thousands,
thousands of thousands ... that he defined. But the
form of the numbers used by al-Khorezmi is unknown,
perhaps they were the letters of the Arabic alphabet or
the Arabic numerals of the East. In fact, a purely
alphabetic number system existed for a very long time,
as evidenced by the "Book of arithmetic for scribes and
traders", written by Abu-l-Wafa between 961 and 976,
and the famous "A sufficient book on the science of
arithmetic," written by al-Karadzhi at the end of the X -
beginning of the XI century.
[1-3]
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Al-Karaji (late X - early XI centuries), a native of the city
of Karaj, located between Tehran and Qazvin, is the
author of many very important works, namely, "A
Sufficient Book on the Science of Arithmetic", "Al-
Fakhri", an extensive algebraic treatise, dedicated to
the vizier of Baghdad Fakhr al-Mulk, as well as the book
"Al-Badi" devoted to the study of indefinite equations.
The Sufficient Book is a textbook of practical arithmetic,
much like another book he wrote between 961 and
976. Abu-l-Wafa and which bears the name "Book of
arithmetic for scribes and merchants". The numbers
were written there verbally and nowhere was the
decimal positional number system used, which was
more in line with the habits of traders. Abu-l-Wafa
considered the theory of fractions in detail. Al-Karaji
also paid attention to the decomposition of ordinary
fractions into the sum of aliquot fractions. Note here
that at the end of the 10th century, arithmetic
algorithms, in particular, algorithms for extracting
square and even cube roots, were significantly
developed. Al-Uqlidizi (about 952-953) gave an
approximation with a lack of square root of the
expression
(2
1)
r
N
a
a
= +
+
.
Other
mathematicians, such as Kushiyar ibn Labban and his
student An-Nasawi, improved these results and
extended them to the cube root using the decimal
representation of the number
1
0
10
...
m
m
N
n
n
−
=
+ +
and the decomposition of the binomial
3
(
)
a b
+
as well
as
3
(
... )
a b
k
+ +
.
[3]
The successful development of arithmetic algorithms
led Al-Karaji and his followers to search for similar
procedures in the case of algebraic expressions. In
addition to the practical part, the "Sufficient Book"
contains the main algebraic part devoted to solving six
canonical types of equations. But his presentation is an
achievement from the point of view of methodology,
because al-Karaji grouped before each problem the
elements of algebraic calculus that are essential for its
solution (transformation of irrational quantities,
identities, etc.). This theoretical orientation was clearly
established in the algebraic treatise "Al-Fakhri". In his
preface, al-Karaji defined the goal of the science of
calculus as determining indefinite quantities using
known ones. It is necessary to turn to the means of
arithmetic calculus and apply them to all expressions
containing unknowns. Thus algebra became explicitly
the arithmetic of the unknown. It can be said that here
its subject was defined for the first time, and the al-
Karaji school expanded the range of methods and
algorithms applied to expressions containing the
unknown.
[4-8]
He applied arithmetic operations to monomials, and
then to expressions composed of monomials, that is, to
polynomials, considering addition and subtraction on
equal terms. As for division, he limited himself to
division into monomials. We get acquainted with the
results of the al-Karaji school concerning division and
square root extraction from the work of his follower al-
Samawal, who continued his research.
But al-Karaji has already managed to describe what
could now be called the algebra of polynomials. These
methods of "arithmetization of algebra", according to
Rashed, are based, on the one hand, on the initial
elements of the algebra of al-Khorezmi and Abu-Kamil,
and on the other hand, on the translation of
Diophantus, performed by Costa ibn Luke under the
title "The Art of Algebra". Indeed, although
"Arithmetic" considered arithmetic on the set of
positive rational numbers, Diophantus used methods
of an algebraic nature in it. These methods influenced
the methods of the Arab algebraists of the second
period, who mastered and developed them. Al-Karaji
summed up many finite arithmetic series, such as for
which he gave a beautiful proof, both geometric and
algebraic. In the text of al-Samawal, which he,
however, attributed to al-Karaj, there is a table of
coefficients for
(
)
n
a b
+
to
12
n
=
, and the author
adds that it can be continued indefinitely in accordance
with the rule of formation, which is now written as
(
1)
(
1)
(
1)
m
m
m
n
n
n
C
C
C
−
−
−
=
+
(the so-called Pascal's triangle ).
Karaji defined algebra as "a method of calculus that
allows, using known quantities, to find unknowns." It
should be considered his merit that he began to solve
algebraic problems exclusively by mathematical
methods, without resorting to geometric ones. Thus,
Karaji is the founder of the algebraic method without
the use of geometric schemes. Thus he showed that
algebra itself is a self-sufficient discipline.
Historian, orientalist and mathematician Franz Wöpke
wrote about al-Karaji: "This is the first mathematician
who proposed the most advanced algebraic theories of
calculus in the Islamic world." Al-Karaji created a table
of binomial coefficients, the principle of their additive
generation and a binomial formula. Al-Karaji was the
first to systematically use algebraic methods of
calculus, worked with definite and indefinite equations,
derived not only square, but also cube roots.
[9-11]
He took algebra beyond the bounds of Euclidean
geometry and contributed to its formation as a
separate discipline. Also, thanks to his work, algebra
began to take the form it has now. In his "Book of
Algebra and Muqabal" the scientist leads to determine
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American Journal of Applied Science and Technology (ISSN: 2771-2745)
the sum of an arithmetic progression, as well as the
sum of squares and cubes of consecutive numbers.
Al-Karaji was not only a mathematician, but also
a hydrologist. In his book On Finding Hidden Waters, he
describes the physical characteristics and vegetation of
the soils under which the water sources are located. He
also gives definitions of water sources and gives
groundwater extraction techniques. All this allows us to
call him the first hydrological engineer.
METHODOLOGY
Al-Karaji systematically studied the algebra of
indicators, and was the first to understand that the
2
3
,
,
,
x x
x
sequence could be extended indefinitely;
and backward
2
3
1
1
1
,
,
,
x
x
x
. However, since, for
example, the product of a square and a cube will be
expressed in words, and not in numbers, like a square-
cube, the numerical property of adding indicators was
not clear.
Al-Karaji gave the first formulation of binomial
coefficients and the first description of Pascal's
triangle. He is also credited with discovering the
binomial theorem. Another important idea introduced
by al-Karaji and continued by al-Samaw'al and another
was that of an inductive argument for solving certain
arithmetic sequences. Thus, al-Karaji used such an
argument to prove a result about the sums of integral
cubes already known. Al-Karaji did not, however, state
the general result for an arbitrary n. He stated his
theorem for a specific integer
10
n
=
. His proof,
however, was clearly intended to be extensible to any
other integer. Al-Karaj's argument includes essentially
two main components of the modern argument by
induction, namely the truth of the statement for
(
)
3
1 1 1
n
=
=
and the originating truths for
n
k
=
from
which of
1
n
k
= −
. Of course, this second component
is not explicit, since, in a sense, al-Karaj's argument is in
the opposite direction; this, it starts at
10
n
=
and goes
down to
1
n
=
, not going up. However, his argument in
al-Fakhri is the earliest surviving proof of the addition
formula for integral cubes.
[12-14]
Al-Karaji operates not only with square but also cube
roots, using the formula for the cube of the sum and
difference. He gives rules for determining the sum of an
arithmetic progression, as well as the sum of squares
and cubes of consecutive numbers. For the sum of
squares, al-Karaji gives the correct formula, but says
that he cannot prove it correct. For the sum of cubes,
he gives a geometric proof. Al-Karaji gives in his essay a
table of binomial coefficients, the principle of their
additive generation and the binomial formula.
[15-21]
Statement of Problem
How are the sums calculated?
If the word is about the sum of two or three people, it
is understandable. But sometimes there are problems,
such as calculating the sum of the terms of a sequence
of large numbers with some connection. This is no
longer an easy task. The problem of calculating a
number of complex sums from the history of
mathematics to the present has attracted the attention
of many mathematicians. Examples include the
manuscripts and historical problems that have come
down to us. Here is one such issue.
Question: If the clock rings 1 time at 1, 2 times at 2, 3
times at 3, and so on, how many times a day will it ring?
In order to find the number of bells, just find the sum
of the numbers from 1 to 12 and multiply by 2. These
numbers are part of the arithmetic progression, so it's
easier to find the sum. If we change the problem as
follows, that is, if the clock rings 1 time when it is 1, 4
times when it is 2, in short, what is the square of the
clock, how many times does it ring in a day? To solve
this problem
12
2
2
2
2
2
1
1
2
3
... 12
k
k
=
= +
+ + +
You need to calculate the sum of the views and multiply
it by 2. We can say that this is not easy anymore. The
formulas for calculating such sums have been used
since ancient times. In this article, we will focus on two
methods of calculating such sums.
Using Newton's binomial formula for calculating
sums
Example 1.
Evaluate
1
1
1 2 3 ...
n
n
k
S
k
n
=
=
= + + + +
.
We use Binomial theorem in order to solve the problem. Namely, From
2
2
2
(
1)
2
1 1
k
k
k
+
=
+ +
We put
1, 2,...,
n
instead of
k
respectively and we rewrite following equalities
2
2
2
,
2
1
2 1 1 1
= + +
2
2
2
,
3
2
2 2 1 1
=
+ +
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2
2
2
,
4
3
2 3 1 1
= + +
.…………………..
2
2
2
.
(
1)
2
1 1
n
n
n
+
=
+ +
We sum all equations and obtain following equalities.
(
)
2
2
2
2
2
2
2
2
2
3
4
(
1)
1
2
3
1 2 ...
,
2
n
n
n
n
+ +
++ +
= +
+ ++
+
+
+ +
+
2
1
2
(
1)
1
2
.
n
n
S
n
+
− =
+
Since
2
2
1
(
1)
2
n
n
n
+
=
+ +
the last equality rewrites as follows
(
)
1
1
.
2
n
n
n
S
+
=
(1)
Example 2.
Evaluate
2
2
2
2
2
2
1
1
2
3
...
n
n
k
S
k
n
=
=
= +
+ + +
.
In this example we use Binomial theorem in order to solve the problem. Namely,
3
3
2
2
3
(
1)
3
1 3
1
1
k
k
k
k
+
=
+
+ +
We put
1, 2,...,
n
instead of
k
respectively and we rewrite following equalities
3
3
2
2
3
,
2
1
3 1 1 3 1 1
1
= +
+ +
3
3
2
2
3
,
3
2
3 2 1 3 2 1
1
= +
+ +
3
3
2
2
3
,
4
3
3 3 1 3 3 1
1
= +
+ +
…………………………….
3
3
2
2
3
.
(
1)
3
1 3
1
1
n
n
n
n
+
=
+ + +
We sum all equations and obtain following equalities.
(
)
(
)
3
3
3
3
3
3
3
3
2
2
2
2
3
4
(
1)
1
2
3
3 1
2
3 1 2 ...
,
n
n
n
n
n
+ + ++ +
= + + ++
+
+
++
+ + + +
+
3
3
2
1
(
1)
1
3
3
.
n
n
n
S
S
n
+
− =
+ +
(2)
From (1) we get
(
)
1
1
2
n
n
n
S
+
=
. We put
(
)
1
2
n
n
+
to (2) and find
2
n
S
(
)
2
3
3
1
3
(
1)
1
3
,
2
n
n
n
S
n
n
+
=
+
− −
−
3
2
2
2
3
3
,
2
n
n
n
n
S
+
+
=
2
(
1)(2
1)
.
6
n
n n
n
S
+
+
=
(3)
Hence, one gets
2
1
(
1)(2
1)
6
n
k
n n
n
k
=
+
+
=
. Now, we can solve the above problem of clock by using formula (3).
For
12
n
=
we get
(
)
12
2
2
2
2
2
1
12 12 1 (2 12 1)
1
2
3
... 12
650
6
k
k
=
+
+
= +
+ + +
=
=
Since the number of all clock bells in 24 hours, obtained result is multiplied by 2. Hence, in all 1300 clock bells.
Example 3.
Evaluate the sum
3
3
3
3
3
3
1
1
2
3
...
n
n
k
S
k
n
=
=
= +
+ + +
.
We use the same method as above. One uses the following equality
4
4
3
2
2
3
4
(
1)
4
1 6
1
4
1
1
k
k
k
k
k
+
=
+ + + +
.
Namely,
4
4
3
2
2
3
4
2
,
1
4 1 1 6 1 1
4 1 1
1
= + + + +
4
4
3
2
2
3
4
3
,
2
4 2 1 6 2 1
4 2 1
1
=
+ + + +
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American Journal of Applied Science and Technology (ISSN: 2771-2745)
4
4
3
2
2
3
4
4
,
3
4 3 1 6 3 1
4 3 1
1
= + + + +
……………………………………
4
4
3
2
2
3
4
(
.
1)
4
1 6
1
4
1
1
n
n
n
n
n
+
=
+ + + +
We sum all equalities and obtain
4
4
3
2
1
(
1)
1
4
6
4
n
n
n
n
S
S
S
n
+
− =
+
+
+
(4)
We put (1) and (3) to (4) and find
3
n
S
.
(
)(
)
(
)
3
4
4
1 2
1
1
4
(
1)
1
6
4
,
6
2
n
n n
n
n
n
S
n
n
+
+
+
=
+
− −
−
−
3
4
3
2
,
4
2
n
S
n
n
n
=
+
+
2
2
3
(
1)
.
4
n
n n
S
+
=
(5)
Hence,
2
2
3
1
(
1)
4
n
k
n n
k
=
+
=
.
Example 4.
Evaluate
1
1
2
3
...
m
m
m
n
n
k
m
m
m
S
k
n
=
=
=
+
+
+ +
.
For calculating the sum we use Binomial theorem, i.e.,
0
(
)
m
m
k
m k
k
m
k
a b
C a
b
−
=
+
=
.
Here
(
)
1
(
1)
!
k
m
m m
m k
C
k
−
− +
=
binomial coefficient. Then we obtain
1
1
1
2
1
2
1
1
1
1 1
(1 1)
1
1
..
1
. 1
m
m
m
m
m
m
m
C
C
+
+
−
+
+
+
+
−
=
+
+ +
1
1
1
2
1
2
1
1
1
2 1
(2 1)
2
...
2
1
1
m
m
m
m
m
m
m
C
C
+
+
−
+
+
+
+
−
+
+ +
=
…………………………..
1
1
1
2
1
2
1
1
1
1
(
... 1
1
1)
m
m
m
m
m
m
m
n
n
C
n
C
n
+
+
−
+
+
+
+
−
=
+
+ +
We sum all equalities and obtain.
1
1
1
2
1
3
2
1
1
1
(
1)
1
...
m
m
m
m
m
m
n
m
n
m
n
n
C
S
S
S
C
C
n
+
+
−
−
+
+
+
+
−
=
+
+
+
(6)
From (6) we find
0
n
A
. Clearly, if
1
k
m
+
, then
1
0
k
m
C
+
=
. Also, if
0
m
=
then (6) can be written as
1
1
1
0
1
(
1)
1
n
n
C S
+
− =
. Consequently, we get
0
n
S
n
=
. If we calculate of (6) for the case
1
m
=
, then one gets
2
2
1
1
2
0
2
2
(
1)
1
n
n
n
C S
C S
+
− =
+
1
(
1
2
)
n
n n
S
+
=
Namely, we prove first part of first example.
Now we put 2 instead of
m
in (6) and find
2
n
S
.
3
3
1
2
2
1
3
0
3
3
3
(
1)
1
n
n
n
n
C
C
S
S
S
C
+
− =
+
+
By using the values of
0
n
S
and
1
n
S
we find the value of
2
n
S
.
2
(
1)(2
1)
.
6
n
n n
n
S
+
+
=
Hence, corollary in the second problem is the same as (2).
By using the process, we can calculate
(
)
m
n
S
m
N
.
For instance, from
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American Journal of Applied Science and Technology (ISSN: 2771-2745)
2
4
4
4
4
4
4
1
(
1)(2
1)(3
3
1)
1
2
3
...
30
n
n
k
n n
n
n
n
S
k
n
=
+
+
+
−
=
= +
+ + +
=
We put necessary equations to (6).
Now we calculate the above summations with other methods. This method is called al-Karaji method or method
of rectangle.
4. Al-Karaji method
At first, we begin calculation from simple summations.
Example 5.
Calculate the sum
1
1
1 2 3 ...
n
n
k
S
k
n
=
=
= + + + +
.
In example 1, we have previously proved that this sum is equal to (1) using Newton's binomial formula. Now we're
going to get the sum by using a rectangular way. Now we're going to get this sum in a rectangular way. To calculate
this sum, we get a square of size
n n
(Figure 1) consisting of
n
columns and rows. Obviously, inside the square,
small squares are formed by the intersection of rows and columns. Then the total number of squares is equal to
2
n n
n
=
.
Figure 1
Figure 2
Now, we separate one row from the top of the square and one column from the right (Figure 2) and calculate the
number of all squares inside the obtained domain.
1 2
1
n n
n
+ − =
−
Hence, we have
2
1
n
−
squares in our first domain.
Similarly, we separate the same domain from other parts of the square (Figure 2) and calculate the number of
squares in it.
(
1)
(
1) 1
2(
1) 1
n
n
n
− +
− − =
− −
Continuing this process to the end, we write the obtained results. It should be noted that each time separated
domains are called al-Karaji gnomons.
2
1,
2(
1) 1, ...,
2 1 1
n
n
−
− −
−
The sum of all these values is equal to the total number of squares in the square in Figure 1.
2
2
2(
1) ... 2 1
n
n
n
n
+
− + + − =
2
2(1 2 ...
)
n
n
n
+ + + − =
(
1)
1 2 ...
2
n n
n
+
+ + + =
Hence,
1
1
(
1)
2
n
n
k
n n
S
k
=
+
=
=
.
Example 6.
Calculate the sum
3
3
3
3
3
3
1
1
2
3
...
n
n
k
S
k
n
=
=
= +
+ + +
.
This example was first considered by the Iranian mathematician Al-Karaji, using the rectangular method described
above. This is why this method is called Al-Karaji method. For solving the above example, Al-Karaji took a square
consisting of
(1 2 3 ...
)
n
+ + + +
columns and rows on each side. (Figure 3)
The total number of squares is equal to
(
)(
)
2
(
1)
1 2 3 ...
1 2 3 ...
(
)
2
n n
n
n
+
+ + + +
+ + + +
=
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Figure 3
Figure 4
Now separate n rows and n columns and calculate the number of squares in it. (Figure 4)
(
)
(
)
2
2
3
2
2
3
1
2
1 2 ...
2
2
n n
n
n
n
n
n
n
n
n
n
+
+ + +
−
=
−
=
+
−
=
There are
3
n
squares in our separated domain.
In the same way, we calculate the number of squares by separating
(
1)
n
−
rows and
(
1)
n
−
columns from the
remaining domain.
(
)
(
)
(
)
(
)
(
) (
) ( )
2
2
3
1
2
1 1 2
1
1
2
1
1
(
1)
2
n n
n
n
n
n
n
n
−
−
+ ++
−
−
−
=
−
−
−
=
−
There are
3
(
1)
n
−
squares in this domain.
Continuing this process to the end and recording all the results. Then the following sequence of values is formed
3
3
3
3
3
, (
1) , (
2) ,
, 2 , 1
n
n
n
−
−
.
The sum of all these values is equal to the total number of squares in Figure 3.
2
2
3
3
3
3
2
(
1)
(
1)
1
2
3
...
(
)
2
4
n n
n n
n
+
+
+
+ + +
=
=
Hence,
2
2
3
3
1
(
1)
4
n
n
k
n n
S
k
=
+
=
=
(see [3]) . We also solved this example with the short multiplication method
above and got the same result (5) as the current one.
Al-Karaji calculated only
3
n
S
with this method. Is it possible to calculate
4
5
6
, , ,
,
, (
1, 2,3, 4,
)
m
n
n
n
n
S S S
S
m
=
using this method as well? We have calculated above the cases
1
m
=
and
3
m
=
.
Throughout the article, we show that all
4
5
6
, , ,
,
, (
2,3, 4,
)
m
n
n
n
n
S S S
S
m
=
's can be calculated using the
rectangular method. This is the novelty part of the article.
Example 7.
Calculate the sum
4
4
4
4
4
4
1
1
2
3
...
n
n
k
S
k
n
=
=
= +
+ + +
.
For this example, we get a rectangle with
2
2
2
2
(1
2
3
...
)
n
+ + + +
row of length and
(1 2 3 ...
)
n
+ + + +
columns
of width. (Figure 5) In a rectangle, a total of square is
(
)
2
2
2
2
(
1)(2
1)
(
1)
(1
2
3
...
) 1 2 3 ...
.
6
2
n n
n
n n
n
n
+
+
+
+
+
+ +
+ + + +
=
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Figure 5
Figure 6
Now separate
n
columns from
2
n
row and calculate the number of squares in it (Figure 6).
(
)
(
)
(
)
(
)(
)
2
2
2
2
3
2
3
4
2
1
1 2
1
5
1
1 2 ...
1
2
...
2
6
6
6
n n
n n
n
n
n
n
n
n
n
n
n
n
n
+
+
+
+ + +
+
+
+ +
−
+
−
=
+
=
It turns out that there is
4
2
5
1
6
6
n
n
+
squares in our area before separation. From the part of the rectangle outside
the domain we have separated, we separate
2
(
1)
n
−
rows and
(
1)
n
−
columns and count the number of squares
in it.
(
) (
)
(
)
(
)
(
)
(
) (
) ( ) ( )(
) ( )
2
3
2
2
2
2
3
4
2
(
1) 1 2 ... (
1)
1 1
2
...
1
1
1
1 2
1
5
1
1
1
1
(
1)
(
1)
2
6
6
6
n
n
n
n
n
n n
n n
n
n
n
n
n
n
−
+ + + −
+
−
+
+ +
−
− −
=
−
−
−
=
−
+
−
− −
=
−
+
−
Hence, there is
4
2
5
1
(
1)
(
1)
6
6
n
n
−
+
−
squares in this domain.
In the same way, we can count to the end and record the results.
4
2
4
2
4
2
5
1
5
1
5
1
(
1)
(
,
, ...,
1)
1
1
6
6
6
.
6
6
6
n
n
n
n
+
−
+
−
+
The sum of all these values is equal to the total number of squares inside the rectangle. Namely
4
4
4
4
2
2
2
2
5
1
(
1)
(
1)(2
1)
(1
2
3
...
)
(1
2
3
...
)
6
6
2
6
n n
n n
n
n
n
+
+
+
+
+
+ +
+
+
+
+ +
=
(
)(
) (
)
4
1 2
1
1
5
1
6
6
2
6
n
n n
n
n n
S
+
+
+
=
−
(
)(
)
2
4
1 2
1 (3
3
1)
30
n
n n
n
n
n
S
+
+
+
−
=
Hence,
(
)(
)
2
4
4
1
1 2
1 (3
3
1)
30
n
n
k
n n
n
n
n
S
k
=
+
+
+
−
=
=
.
Example 8.
Calculate the sum
5
5
5
5
5
5
1
1
2
3
...
n
n
k
S
k
n
=
=
= +
+ + +
.
To solve the example, we consider a square shape whose width and height are equal, that is, both have
2
2
2
2
(1
2
3
...
)
n
+ + + +
rows and columns. (Figure 7)
Figure 7
Figure 8
The total number of squares inside the square is
2
2
2
2
2
2
2 2
(
1) (2
1)
(1
2
3
...
)
36
n n
n
n
+
+
+
+ + +
=
(7)
Now, we count the number of squares on each side of the square by separating
2
n
columns and rows. (Figure 8)
(
)(
)
2
2
2
2
2
4
2
4
5
3
1 2
1
2
1
2
(1
2
3
...
)
2
6
3
3
n n
n
n
n
n
n
n
n
n
+
+
+
+ + +
−
=
−
=
+
By separating
2
(
1)
n
−
more such rows and columns from the rest of the square (Figure 8), we calculate the
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number of squares in it.
(
)
(
) (
)(
) ( )
4
2
2
2
2
2
2
4
5
3
2(
1) (1
2
3
... (
1) )
1
1 2
1
2
1
2
1
1
(
1)
(
1)
6
3
3
n
n
n
n n
n
n
n
n
n
−
+
+ + + −
−
−
=
−
−
=
−
−
−
=
−
+
−
Continuing this work to the end, we record the results obtained.
5
3
5
3
5
3
2
1
2
1
2
1
(
1)
(
,
, ...,
1)
1
1
3
3
3
.
3
3
3
n
n
n
n
+
−
+
−
+
Since these values represent the number of squares in the fields we have separated at each step, their sum is
equal to the value given in (7) for the total number of squares inside the square in Figure 7.
(
)
2
2
2
5
5
5
5
3
3
3
3
2
1
(
1) (2
1)
(1
2
3
...
)
1
2
3
...
3
3
36
n n
n
n
n
+
+
+
+ + +
+
+
+ + +
=
2
2
2
5
3
2
1
(
1) (2
1)
3
3
36
n
n
n n
n
S
S
+
+
+
=
2
2
2
2
2
5
(
1) (2
1)
(
1)
2
12
4
n
n n
n
n n
S
+
+
+
=
−
2
2
2
5
(
1) (2
2
1)
12
n
n n
n
n
S
+
+
−
=
Hence,
2
2
2
5
5
5
5
5
5
1
(
1) (2
2
1)
1
2
3
...
12
n
n
k
n n
n
n
S
k
n
=
+
+
−
=
= + + + +
=
.
It can be concluded from the above examples. You can calculate any sum
m
n
S
(
1, 2, 3,
)
m
=
using the
rectangular method. To do this, if
2
m
r
=
(
)
r
N
, it is enough to take the height
1
2
3
...
r
r
r
r
n
+
+ + +
and
1
1
1
1
1
2
3
...
r
r
r
r
n
−
−
−
−
+
+
+ +
width of a rectangle (Figure 9).
Figure 9
Figure 10
If
2
1
m
r
=
+
(
0,1, 2, 3,...)
r
=
, it is sufficient to obtain a rectangle in the form of a square with
1
2
3
...
r
r
r
r
n
+
+ + +
sides on both sides (Figure 10). The calculation process is similar to the examples above.
For example, it is easy to prove that the sum of
2
2
2
2
2
2
1
1
2
3
...
n
n
k
S
k
n
=
=
= +
+ + +
is equal to
(
)(
)
1 2
1
6
n n
n
+
+
.
CONCLUSION
The introductory part of the article gives a brief account
of the great Iranian mathematician al-Karaji, his
contribution to the science of algebra, and the opinions
left about him by historians.
The research methodology section discusses the areas
of algebra studied by al-Karaji, his achievements,
experiences, and innovations in this area. It contains
the specific methods and results he used in the
calculation of some sets, the characteristics of these
sets, and evidence of the role of these works in the
history of mathematics.
The problem-solving part of the article deals with the
general methods of solving the sums formed by any
degree of the sequence of natural numbers. First, the
method of calculation using the well-known method,
American Journal of Applied Science and Technology
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American Journal of Applied Science and Technology (ISSN: 2771-2745)
the formula of short multiplication, is described. Using
this method, several sums were calculated and the
general result was given.
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