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volume 4, issue 5, 2025
201
ADVANCED APPROACHES TO SUMMATION IN MATHEMATICAL EDUCATION
Denov Institute of Entrepreneurship and Pedagogy
Abdurashidov Nuriddin
abdurashidovnuriddin9550@gmail.com
Students of the Denov Institute of Entrepreneurship and Pedagogy
Joʻrayeva Sevinch
Ibadullayev Diyorbek
Abstract
: This article explores the theoretical and practical aspects of efficiently and optimally
calculating certain types of mathematical sums that are widely used in mathematical analysis and
algebraic structures. Since secondary school mathematics textbooks typically address only basic
summation cases using arithmetic or geometric progression formulas, students are often limited
to solving simple summation problems. As a result, they face difficulties when encountering
more complex expressions. In this study, general formulas for various types of sums, their
mathematical proofs, conditions of applicability, and illustrative examples are thoroughly
discussed. In addition, several problems related to identities and summations frequently
encountered in contemporary mathematical olympiads and academic competitions are analyzed.
This work serves as a theoretical and methodological resource for teaching mathematics,
developing computational algorithms, and enhancing students’ mathematical thinking.[1-5].
Keyword
:Mathematics education,Sums,Sum calculation methods, Advanced approache,
Innovative teaching, Algebraic sums.
1-M.
Calculate
the
sum:
3
2
1
ta
...
...
n
a
aaa
aaa
aa
a
S
+
+
+
+
=
Solution.
To evaluate this sum, we transform the given expression into the following form:
(
)
(
)
(
)
(
)
(
)
81
9
10
10
9
1
10
10
9
1
10
1
10
10
9
10
...
10
10
10
9
1
10
...
1
10
1
10
1
10
9
9
...
999
...
999
99
9
9
...
...
1
3
2
3
2
ta
ta
n
a
n
a
n
a
n
a
a
a
a
aaa
aaa
aa
a
S
n
n
n
n
n
n
n
-
-
=
-
-
=
-
-
-
=
=
-
+
+
+
+
=
-
+
+
-
+
-
+
-
=
=
+
+
+
+
=
+
+
+
+
=
+
3
2
1
3
2
1
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202
As a result,
3
2
1
ta
...
...
n
a
aaa
aaa
aa
a
S
+
+
+
+
=
we find
(
)
81
9
10
10
1
n
a
S
n
-
-
=
+
that the given
sum can be evaluated using a known.[5-9].
2-M.
n
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
...
4
3
2
1
1
...
4
3
2
1
1
3
2
1
1
2
1
1
evaluate
the
sum.
Solution.
It is evident that the denominators form the sum of the first nnn terms of an arithmetic
progression. Therefore,
(
)
1
1
1
2
1
1
1
2
1
2
1
1
1
...
5
1
4
1
4
1
3
1
3
1
2
1
2
1
2
...
5
4
2
4
3
2
3
2
2
2
1
1
...
4
2
4
1
1
3
2
3
1
1
2
2
2
1
1
...
4
3
2
1
1
...
4
3
2
1
1
3
2
1
1
2
1
1
+
-
=
+
-
+
=
+
-
=
=
+
-
+
+
-
+
-
+
-
=
+
+
+
+
+
=
+
+
+
+
+
+
+
+
+
=
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
n
n
n
n
n
n
n
n
n
n
n
n
We obtain the result. Thus, the value of the given sum
1
1
+
-
n
n
i s equal.[10-13]
========================================================
3-M. Calculate
:
....
3
5
1
2
4
1
1
3
1
2
2
2
+
+
+
+
+
+
=
S
Solution.
First, we write down the general term formula and then perform the corresponding
calculations
(
)
+
-
+
=
+
+
=
+
+
=
4
1
1
1
3
1
4
5
1
2
1
2
2
n
n
n
n
n
n
a
n
We find its value. Now, we
apply this formula to the given sum
36
13
4
1
3
1
2
1
3
1
....
9
1
6
1
8
1
5
1
7
1
4
1
6
1
3
1
5
1
2
1
3
1
....
3
5
1
2
4
1
1
3
1
2
2
2
=
+
+
=
=
+
-
+
-
+
-
+
-
+
-
=
+
+
+
+
+
+
=
S
4-M. Calculate the sum:
100
99
98
1
...
5
4
3
1
4
3
2
1
3
2
1
1
+
+
+
+
=
S
Solution.
We write down the general term formula and then perform the corresponding
calculations,
(
)(
)
(
)(
)
(
)
(
)(
)
(
) (
)(
)
+
+
-
+
=
+
+
-
+
=
+
+
=
+
+
=
2
1
1
1
1
2
1
2
1
2
2
1
2
1
2
2
1
2
1
1
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
a
n
(
) (
)(
)
+
-
+
-
+
-
=
+
+
-
+
=
2
1
1
1
1
1
1
2
1
2
1
1
1
1
2
1
n
n
n
n
n
n
n
n
a
n
We
find
its
value.
Now,
we
apply
this
formula
to
the
given
sum,
-
=
-
-
-
=
+
-
+
-
+
-
=
=
=
100
99
1
2
1
2
1
100
1
2
1
99
1
1
1
2
1
2
1
1
1
1
1
1
2
1
1
1
n
k
n
k
k
n
n
n
n
a
.[13
-18].
5-M. Calculate the sum:
...
5
...
5
3
5
2
5
1
3
2
+
+
+
+
+
=
n
n
S
Solution.
We multiply both parts of the given sum by 5. Then,
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203
...
5
...
5
3
5
2
1
5
1
2
+
+
+
+
+
=
-
n
n
S
We form an equation. Subtracting the given equation from
this
4
5
5
1
1
1
...
5
1
...
5
1
5
1
1
...
5
1
5
...
5
2
5
3
5
1
5
2
1
4
1
2
1
1
2
2
=
-
=
+
+
+
+
+
=
=
+
-
-
+
+
-
+
-
+
=
-
-
-
n
n
n
n
n
S
6-M. Calculate the sum:
...
4
3
2
4
3
2
+
+
+
+
=
x
x
x
x
S
, in this,
1
<
x
.
Solution.
We divide the
...
4
3
2
4
3
2
+
+
+
+
=
x
x
x
x
S
given sum by x. As a result
Consequently
...
4
3
2
1
3
2
+
+
+
+
=
x
x
x
x
S
An
equation
is
formed.
Now,
...
1
3
2
+
+
+
+
=
-
x
x
x
S
x
S
We examine the difference. It is evident that this corresponds to
the sum of an infinitely decreasing geometric progression. Therefore,
x
S
x
S
-
=
-
1
1
It holds.
Simplifying the final equation,
(
)
2
1
x
x
S
-
=
we obtain the result.[18-21].
7-M. Calculate the sum:
!
1000
999
....
!
4
3
!
3
2
!
2
1
+
+
+
+
Solution.
If we express the given sum in the following form and perform the necessary
simplifications:
!
1000
1
1
!
1000
1
!
999
1
.....
!
4
1
!
3
1
!
3
1
!
2
1
!
2
1
!1
1
!
1000
1
1000
...
!
4
1
4
!
3
1
3
!
2
1
2
!
1000
999
....
!
4
3
!
3
2
!
2
1
-
=
-
+
+
+
-
+
-
+
-
=
-
+
+
-
+
-
+
-
=
+
+
+
+
we obtain the result.
8-M.
1
3
2
...
4
3
2
1
-
+
+
+
+
+
=
n
nx
x
x
x
S
calculate the sum.
Solution. 1-usul.
To compute the given sum conveniently, we first consider it as a function and
find its (denoted as Q) antiderivative. Then,
(
)
1
1
...
3
2
-
-
=
+
+
+
+
=
x
x
x
x
x
x
x
Q
n
n
It works.
Here, we used the formula for the sum of the first n terms of a geometric progression. Now, if we
differentiate the final formula we obtained, we get the value of the sum
S
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2
2
1
1
1
1
1
1
1
1
1
1
-
+
-
-
=
-
-
-
-
-
+
=
-
-
=
=
+
+
x
x
n
nx
x
x
x
x
x
n
x
x
x
Q
S
n
n
n
n
.
That is to say
(
)
(
)
2
1
3
2
1
1
1
...
4
3
2
1
-
+
-
-
=
+
+
+
+
+
=
-
x
x
n
nx
nx
x
x
x
S
n
n
will be.
2-usul.
If we multiply both parts of the given sum by xxx, then
n
nx
x
x
x
x
Sx
+
+
+
+
+
=
...
4
3
2
4
3
2
We get the result. If we subtract this sum from the given
equation,
n
n
n
n
nx
x
x
nx
x
x
x
x
Sx
S
-
-
-
=
-
+
+
+
+
+
=
-
-
1
1
...
1
1
3
2
It works. From this,
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204
then
(
)
(
)
(
)
(
)
2
2
1
1
1
1
1
)
1
(
S
1
1
1
-
+
-
-
=
-
+
-
-
=
-
-
-
=
-
-
x
x
n
nx
x
x
x
nx
nx
x
x
x
S
n
n
n
n
n
results
in.[22-25].
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