The document discusses mathematical concepts related to telescoping series. It provides 30 problems involving techniques like factorization, summation of series, and algebraic manipulation for preparing a mathematics competition. The problems cover topics such as factoring expressions, evaluating infinite series, and simplifying fractions. A disclaimer is also included, stating that the material was prepared by Tulus Budi Prasetyo.

1. KONSEP DASAR
PEMBAHASAN SOAL

2. 2
• Not all series are telescoping.
• Don’t look for them everywhere!
• But don’t miss them when you see them
DISCLAIMER:
Tulus Budi Prasetyo, S.Si

3. 3
PROBLEM
1
PROBLEM
2
PROBLEM
3
PROBLEM
4
Materi Prinsip dan Teknik Teleskoping
PERSIAPAN KOMPETISI MATEMATIKA
1
1 × 2
+
1
2 × 3
+
1
3 × 4
+ ⋯ +
1
9 × 10
=
1
4
+
1
28
+
1
70
+ ⋯ +
1
9700
=
1
1.2.3
+
1
2.3.4
+
1
3.4.5
+ ⋯ +
1
10.11.12
=
1
1.2.3.4
+
1
2.3.4.5
+
1
3.4.5.6
+ ⋯ +
1
10.11.12.13
=
1
𝑛 × (𝑛 + 1)
=
1
𝑛
−
1
𝑛 + 1
1
𝑛 × (𝑛 + 𝑘)
=
1
𝑘
1
𝑛
−
1
𝑛 + 𝑘
1
𝑛 × (𝑛 + 1) × (𝑛 + 2)
1
𝑛 × (𝑛 + 2)
×
1
(𝑛 + 1)
=
1
(𝑛 + 1)
×
1
2
1
𝑛
−
1
𝑛 + 2
1
𝑛 × (𝑛 + 3)
×
1
(𝑛 + 1)(𝑛 + 2)
=
1
(𝑛 + 1)(𝑛 + 2)
×
1
3
1
𝑛
−
1
𝑛 + 3
1
3
1
𝑛(𝑛 + 1)(𝑛 + 2)
−
1
(𝑛 + 1)(𝑛 + 2)(𝑛 + 3)
PROBLEM
5
3
1
+
3
1 + 2
+
3
1 + 2 + 3
+ ⋯ +
3
1 + 2 + 3 + ⋯ + 100
3
1 + 2 + 3 + ⋯ + 𝑛
=
3
𝑛 × (𝑛 + 1)
2
3
1 + 2 + 3 + ⋯ + 𝑛
=
3 × 2
𝑛 × (𝑛 + 1)
3
1 + 2 + 3 + ⋯ + 𝑛
=
6
𝑛
−
6
𝑛 + 1

4. 4
Materi Prinsip dan Teknik Teleskoping
PERSIAPAN KOMPETISI MATEMATIKA
6
7
8
9
10
11
12
13
14
15
16

5. 5
Materi Prinsip dan Teknik Teleskoping
PERSIAPAN KOMPETISI MATEMATIKA
17
18
19
20
21
23
24

6. 6
PROBLEM
25
Materi Prinsip dan Teknik Teleskoping
PERSIAPAN KOMPETISI MATEMATIKA
Misalkan 𝑎, 𝑏, 𝑐, 𝑑𝑎𝑛 𝑑 adalah konstanta sedemikian sehingga ekspresi berikut berlaku untuk
setiap bilangan real yang tidak mengakibatkan penyebut bernilai nol.
Berapakah nilai dari : 𝑎 + 𝑏 + 𝑐 + 𝑑
PROBLEM
26
PROBLEM
27
PROBLEM
28
PROBLEM
29
PROBLEM
30

7. 7

8. 8
PROBLEM
1
PROBLEM
2
1
1 × 2
+
1
2 × 3
+
1
3 × 4
+ ⋯ +
1
9 × 10
=
1
4
+
1
28
+
1
70
+ ⋯ +
1
9700
=
1
1 1
2
1
1 × 2
1
2
1
3
1
2 × 3
1
3
1
4
1
3 × 4
1
1
−
1
2
=
2 − 1
1 × 2
=
1
1 × 2
1
2
−
1
3
=
3 − 2
2 × 3
=
1
2 × 3
1
3
−
1
4
=
4 − 3
3 × 4
=
1
3 × 4
1
1
−
1
2
+
1
2
−
1
3
+
1
3
−
1
4
+ ⋯ +
1
9
−
1
10
=
1
1
−
1
10
=
9
10
1
1 × 4
+
1
4 × 7
+
1
7 × 10
+ ⋯ +
1
97 × 100
=
1
3
4 − 1
1 × 4
+
7 − 4
4 × 7
+
10 − 7
7 × 10
+ ⋯ +
100 − 97
97 × 100
=
1
3
1
1
−
1
4
+
1
4
−
1
7
+
1
7
−
1
10
+ ⋯ +
1
97
−
1
100
=
1
3
1
1
−
1
100
=
1
3
99
100
=
33
100
1
𝑛 × (𝑛 + 1) × (𝑛 + 2)
1
𝑛 × (𝑛 + 2)
×
1
(𝑛 + 1)
=
1
(𝑛 + 1)
×
1
2
1
𝑛
−
1
𝑛 + 2
1
2
1
𝑛(𝑛 + 1)
−
1
(𝑛 + 1)(𝑛 + 2)
1
2
1
1.2
−
1
2.3
+
1
2
1
2.3
−
1
3.4
+
1
2
1
3.4
−
1
4.5
+
1
2
1
10.11
−
1
11.12
+ ⋯
65
264
1
2
1
1.2
−
1
11.12
1
2
1
2
−
1
132
1
2
66
132
−
1
132
PROBLEM
3
1
1.2.3
+
1
2.3.4
+
1
3.4.5
+ ⋯ +
1
10.11.12
=

9. 9
PROBLEM
4
1
1.2.3.4
+
1
2.3.4.5
+
1
3.4.5.6
+ ⋯ +
1
10.11.12.13
=
PROBLEM
5
3
1
+
3
1 + 2
+
3
1 + 2 + 3
+ ⋯ +
3
1 + 2 + 3 + ⋯ + 100
1
𝑛 × (𝑛 + 1) × (𝑛 + 2) × (𝑛 + 3)
1
𝑛 × (𝑛 + 3)
×
1
(𝑛 + 1)(𝑛 + 2)
=
1
(𝑛 + 1)(𝑛 + 2)
×
1
3
1
𝑛
−
1
𝑛 + 3
1
3
1
𝑛(𝑛 + 1)(𝑛 + 2)
−
1
(𝑛 + 1)(𝑛 + 2)(𝑛 + 3)
1
3
1
1.2.3
−
1
2.3.4
+
1
3
1
2.3.4
−
1
3.4.5
+
1
3
1
3.4.5
− ⋯ +
1
3
⋯ −
1
11.12.13
1
3
1
1.2.3
−
1
11.12.13
1
18
−
1
3.11.12.13
3
1 + 2 + 3 + ⋯ + 𝑛
=
3
𝑛 × (𝑛 + 1)
2
3
1 + 2 + 3 + ⋯ + 𝑛
=
3 × 2
𝑛 × (𝑛 + 1)
3
1 + 2 + 3 + ⋯ + 𝑛
=
6
𝑛
−
6
𝑛 + 1
3
1
+
6
2
−
6
3
+
6
3
−
6
4
+
6
4
−
6
5
+ ⋯ +
6
100
−
6
101
=
3
1
+
6
2
−
6
101
= 6 −
6
101
= 6 1 −
1
101
= 6
100
101
=
600
101

10. 10
PROBLEM
6
PROBLEM
7
PROBLEM
8

11. 11
PROBLEM
9
PROBLEM
10

12. 12
PROBLEM
11
PROBLEM
12

13. 13
PROBLEM
13
PROBLEM
14

14. 14
PROBLEM
15
PROBLEM
16

15. 15
PROBLEM
17
PROBLEM
18

16. 16
PROBLEM
19

17. 17
PROBLEM
20
PROBLEM
21
PROBLEM
22

18. 18
PROBLEM
23

19. 19
PROBLEM
25
Misalkan 𝑎, 𝑏, 𝑐, 𝑑𝑎𝑛 𝑑 adalah konstanta sedemikian sehingga ekspresi berikut berlaku untuk
setiap bilangan real yang tidak mengakibatkan penyebut bernilai nol.
Berapakah nilai dari : 𝑎 + 𝑏 + 𝑐 + 𝑑
SOLUSI

20. 20
PROBLEM
26
SOLUSI
SOLUSI
PROBLEM
27

21. 21
SOLUSI
PROBLEM
28

22. 22
SOLUSI
PROBLEM
29

23. 23
SOLUSI
PROBLEM
30