B. Abdurahmonov Toshkent–2008 2 B. Abdurahmonov. Matematik induksiya metodi
-§. Natural sonlarning bo’linishini isbotlash
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Matematik induksiya metodi @aniq fan
- Bu sahifa navigatsiya:
- 1. 1-teorema.
- Isbotlash .
- 2. 1-teorema.
- Isbotlash.
- 8.3-masala.
- 1-teorema.
- Quyidagilarni isbotlang
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8-§. Natural sonlarning bo’linishini isbotlash 8.1-masala. n N ∀ ∈
da quidagilarni isbotlang: 1. 2 1 2
2 n n + + + ifoda 7 ga bo’linadi; 2. 3 5
n + ifoda 6 ga bo’linadi. 1. 1-teorema. n = 1 da
3 3 3 2 27 8 35,
+ = + = ga ega bo’lamiz va 7 ga bo’linadi. 2-teorema. 2 1
2 3 2 k k + + + ifodaning 7 ga bo’linishi berilgan. Nujno dokazat’ chto 2 3 3 3 2 k k + + + ifodaning 7 ga bo’linishini isbotlash lozim. Isbotlash. 2( 1) 1 1 2 2 1 2 2 2 3 2 3 9 2 2 9 2
9 2 k k k k k k + +
+ + + + + + + = ⋅ +
⋅ + ⋅ − ⋅
= ( ) 2 1 2 2 9 3
2 2 (2 9) k k k + + + = ⋅
+ + − . Har bir ikkita qo’shiluvchi 7 ga bo’linadi. 2-teorema isbotlandi. 2. 1-teorema. n = 1 da
3 1 5 1 6. + ⋅ = ega bo’lamiz. Bunda 6 soni 6 ga bo’linadi.
3 5 k k + ifodaning 6 ga bo’linishi berilgan. 3 ( 1) 5( 1)
k + + +
ifodaning 6 ga bo’linishini isbotlash lozim. Isbotlash. 3 3 ( 1) 5( 1) 5 3 ( 1) 6 k k k k k k + + + = + + + + . 3 5 k k + yig’indi 6 ga faraz bo’yicha bo’linadi. Tak kak proizvedenie k (
+1) ko’paytma k natural sonda 2 ga bo’linadi, u holda 3 ( 1)
+ ifoda 6 ga bo’linadi. 8.2-masala. n N ∀ ∈
da quyidagilarni isbotlang: 1. 6 15
n ifoda 7 ga bo’linadi; 2. n n − 3 ifoda 3 ga bo’linadi; 1. 1-teorema. n = 1 da
1 15 6 21 + = ega bo’lamiz,ya’ni 21 soni 7 ga bo’linadi.
15 6 k + ifodaning 7 ga bo’linishi berilgan. 1 15 6 k + + ifodaning 7 ga bo’linishini isbotlash lozim. Isbotlash. 1 15 6 15 15 6 k k + + = ⋅ + =
61
( ) ( ) ( ) 7ga bo'linadi 7 ga bo'linadi 15 6 15 15 6 6 15 6 15
15 1 6 k k = + ⋅ − ⋅ + = + ⋅ − − ⋅
. 2-teorema isbotlandi. 2. 1-teorema. n = 1 da
3 1 1 0
− = ga ega bo’lamiz . Bundan 0 soni 3 ga bo’linadi.
3
k − ifodaning 3 ga bo’linishi berilgan. 3 ( 1) ( 1)
k + − + ifodaning 3 ga bo’linishini isbotlash lozim. Isbotlash. 3 3 2 ( 1) ( 1) 3 3 1 1 k k k k k k + − + = + + + − − = ( ) ( ) 3 2 3ga bo'linadi 3ga bo'linadi 3
= − − + ⋅
. 2-teorema isbotlandi. 8.3-masala. Har bir n natural son uchun 7
− ga karrali ekanligini isbotlang. 1-teorema. n = 1 da
7 1 1 0 − = son 7 ga karrali. 2-teorema. n = k da
7 k k − soni 7 ga bo’linishi berilgan. n = k +1 da
7 ( 1) ( 1)
k + − + sonning 7 ga bo’lishini isbotlash losim. Isbotlash. n = 7 satrdan foydalanib, Pascal uchburchgida ( k +1) yig’indining 7-darajasi uchun quyidagilarni hosil qilamiz: 7 7 6 5 4 3 2 ( 1) ( 1) 7 21 35 35 21 7 1 1
k k k k k k k k k + − + = + + + + + + + − − = N 7
5 4 3 2 7 'li 7 'li
7 21 35 35 21 7 ga bo nadi ga bo nadi k k k k k k k k = − + + + + + + . 2-teorema isbotlandi. 8.4-masala. Har bir n natural son uchun 4 15
n + ifodaning 9 ga 1 qoldiqda bo’linishini isbotlang. Isbotlash. Masala sharti 4 15 1 n n + − sonning 9 ga karrali shartiga teng kuchli. 62
1-teorema. n = 1 da
1 4 15 1 1 18 + ⋅ − = ifoda 9 ga karrali. 1-teorema isbotlandi.
da
4 15 1 k k + − sonning 9 ga bo’linishi berilgan. n = k+1
da 1 4 15( 1) 1 k k + + + − sonning 9 ga bo’linishini isbotlash lozim. Isbotlash. ( ) 1 4 15( 1) 1 4 4 15 1 15 1 15( 1) 1
k k k k k k + + + − = + − − + + + − =
( ) ( ) 9 'li 9 'li 4 4 15 1 60 4 15
15 1 4 4 15 1 ( 45 18)
k k ga bo nadi ga bo nadi k k k k k + − − + + + − = + − + −
+
. 9-§. Turli masalalar
9. 1-masala. Tengsizlikni isbotlang 1 1
1 ln(
1) 1 ...
ln 1, , 2 2 3
n n n N n n + < + + + + < + ∀ ∈ ≥
Isbotlash. Dastlab quyidagi tengsizlikni isbotlaymiz: 1 1 1 1 1 1
1 ...
ln 1 ... , , 2 2 3 2 3
1 n dx n n N n n x n + + + <
= < + + + + ∀ ∈
≥ − ∫ . 1 ln n dx n x = ∫ integral 1 ( ) f x x = egri chiziqning [1, n ] oraliqda chegaralangan yuzasiga teng.
63
Bu yiza to’g’ri to’rtburchallar yuzalarining birlashgan yuzasidan kattadir.
1 1 1 ... ln , 2. 2 3 n n N n n + + + <
∀ ∈ ≥ (9.1.1) Bu yuza esa to’rtburchallar yuzasining birlashgan yuzasidan kichikdir.
1 1 1 1 ... ln , , 2 2 3 1
n N n n + + + +
> ∀ ∈
≥ − . (9.1.2) Matematik induksiya metodi bilan (9.1.1) va (9.1.2) tengsizlikni isbotlaymiz: 1-teorema. Uchta rasm va [0, 1] oraliqda aniqlangan integral xossasidan quyidagi tengsizlik kelib chiqadi 2 1
1 ln 2 1 1
2 dx x ⋅ <
= < ⋅ ∫ . Bundan n = 2
da
64
(9.1.1) va (9.1.2) tengsizliklarning o’rinli ekanligi tasdiqlanadi. 1-teorema isbotlandi. 2-teorema. n = k quyidagi tengsizlikning bajarilishi berilgan 1 1 1
1 ...
ln 1 ... , 2 2 3 2 3 1
k k k + + + <
< + + + + ≥ − . n = k+1 da ushbu tengsizlikning bajarilishini isbotlash lozim: 1 1 1
1 1 1 1 ... ln(
1) 1 ...
, 2 2 3 1 2 3
1 k k k k k k + + + +
< + < + + + + + ≥
− . (9.1.3) Isbotlash. Quyidagi shakllarning 1 2 3 , , S S S yuzalarini taqqoslaymiz:
(9.1.3) tengsizlikning chap qismini taqqoslaymiz: = +
⋅ +
< + + + + + ∫ +1 ln 1 1 1 ln 1 1 1 ...
3 1 2 1 k k x x d k k k k k
ln ln( 1) ln
ln( 1),
2 k k k k k = + + − = + ≥ .
(9.1.3) tengsizlikning o’ng qismini taqqoslaymiz: ln(
1) ln ln(
1) ln k k k k + =
+ + −
=
1 1 ln
k k dx k x +
= +
1 1
1 1 1 ... , 2 2 3 1
k k < + + + + + ≥ − .
65
2-teorema isbotlandi. 1 va 2-teoremalarning isbotidan ixtiyoaiy natural 2
≥
9. 2-masala. Ushbu qonuniyatning tarqalish muhiti qanday: 1 1
1 1 ⋅ = ;
1 2 1 22 22 121 + + ⋅ = ;
1 2 3 2 1 333 333 12321
+ + + + ⋅ = ; 1 2 3 4 3 2 1 4444 4444 1234321
+ + + + + + ⋅ =
Gipotezani isbotlaymiz. Har bir
natural son uchun quyidagi tenglik to’g’ri: 2 ...
1 2...( 1) (
1)...2 1 1 2 ... ( 1) (
n nn n n n n n n n ⋅ − − ⋅ =
+ + + − + + − + + +
. (9.2) Matematik induksiya metodi yordamida gipotezani isbotlaymiz. 1-teorema. n = 1 da
1 1 1 1 ⋅ = ga ega bo’lamiz. 1-teorema isbotlandi. (9.2) tenglikning o’ng qismi maxraji yig’indisini topamiz: 1 2 ... ( 1) ( 1) ( 1) ... 2 1 n n n n n + + + − + + + + + − + + + =
2
1) ( 1) 2 2
n n n n ⋅ −
⋅ − = + + = . U holda P P P P
2 ...
... ...
... 1 2 ... ( 1) (
( 1) ... 2 1 n ta n ta n ta n ta nn n nn n nn n nn n n n n n n n ⋅ ⋅ = = + + + − + + + + + − + + + P P
P P ...
... 11...1 11...1 n ta n ta n ta n ta nn n nn n n n = ⋅ = ⋅ . Natijada quyidagi tenglikni isbotlash lozim: P P 12...(
1) ( 1)...21 11...1 11...1 n ta nta n n n − − = ⋅ (9.2.1) 2-teorema. n=k da (9.2.1) tenglikning bajarilishi berilgan. P P 12...(
1) ( 1)...21 11...1 11...1 k ta k ta k k k − − = ⋅ . n = k+1 da (9.2.1) tenglikning bajarilishi ni isbotlash lozim: 66
P P 1 12... ( 1) ...21 11...1 11...1 k ta k ta k k k + + = ⋅ . Isbotlash. 2 2 1 (11...1) (11...1) ((11...1) 10 1) ((11...1) 10) 2 (11...1) 10 1 k ta k ta k ta k ta k ta + ⋅ = ⋅ +
= ⋅ + ⋅ ⋅ + =
1 1 12...( 1) (
1) ...2100 22...20 1 k ta k ta k k k k + + = − − + + =
12...(
1) ( 1) (
1)...21 k k k k k = − + − . 2-teorema isbotlandi. 9.3-masala. Ixtiyoriy juft n natural son uchun tengsizlikni isbotlang: 3 2 1
3 2 1
... sin
... , 0
3! (2 1)! 3! (2 1)! 2 n n x x x x x x x x n n π − + − + − ≤ ≤ −
+ + ≤ ≤
− + . (9.3) Isbotlash. Isbotlashda quyidagi ma’lumotlar talab etiladi: 0, 2 π ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ kesmada sin , cos
funktsiyalar quyidagi shartlarni qanoatlantiradi: a) Ushbu funktsiyalarning yuqori chegarasi o’zgaruvchi bo’lgan integrallar:
0 0 sin
cos , cos sin
x x tdt x tdt x = −
= ∫ ∫ ; (9.3.1) b) 0 sin 1, 0 cos
1 x x ≤ ≤ ≤ ≤ . (9.3.2) 1-teorema. n = 2 da quyidagi tengsizliklarning o’rinli ekanligini isbotlaymiz : 3 3
sin , 0
3! 3! 5! 2 x x x x x x x π − ≤ ≤ −
+ ≤ ≤
0, 2 x π ⎛ ⎞ ∈⎜ ⎟ ⎝ ⎠ da 0 0 sin cos 1
x x tdt dt x = ≤ = ∫ ∫ ga ega bo’lamiz, ya’ni: sin x x ≤ .
ni t ga almashtirish va ushbu tengsizlikni t bo’yicha 0 dan 0, 2
π ⎛ ⎞ ∈⎜ ⎟ ⎝ ⎠ gacha
tekshirganda 2 0 0 0 sin , sin 1 cos
2 x x x x tdt tdt tdt x ≤ = = − ∫ ∫ ∫ ni hosil qilamiz. U holda 67
0, 2 x π ⎛ ⎞ ∈⎜ ⎟ ⎝ ⎠ da 2 cos
1 2
x ≥ −
tengsizlikka teng kuchli 2 1 cos 2 x x − ≤ tengsizlik bajariladi. X ni
T ga almashtiruvchi va ushbu tengsizlikni T bo’yicha 0 dan 0, 2
π ⎛ ⎞ ∈⎜ ⎟ ⎝ ⎠ gacha integrallovchi protseduragani takrorlaymiz: x dt t x x dt t dt t x x x sin
cos , 2 3 2 1 cos 0 3 0 2 0 = ⋅ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ≥ ∫ ∫ ∫ . Shuning uchun
x x sin
! 3 3 ≤ − , ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∈ 2 , 0 π
. (9.3.4) Yana bir bor shuni takrorlaymiz: 3 3
4 0 0 0 sin
1 cos , 3! 3! 2 4!
x x t t x x t dt tdt x t dt ⎛ ⎞ ⎛ ⎞ − ≤ = −
− = − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∫ ∫ ∫ . Oldingiga o’xshash quyidagiga ega bo’lamiz: 2 4 2 4 1 cos cos 1 2 4! 2 4! x x x x x x − ≥ − ⇔ ≤ − + ,
0, 2
π ⎛
∈⎜ ⎟ ⎝ ⎠ . Tortinchi marotaba X ni T ga almashtirish va ushbu tengsizlikni T bo’yicha 0 dan ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∈ 2 , 0 π x gacha integrallab, quyidagiga ega bo’lamiz: ! 5
3 sin
5 3
x x x + − ≤ . (9.3.5) (9.3.4) va i (9.3.5) tengsizliklardan quyidagi hosil bo’ladi: 3 sin 3! x x x − ≤ 3 5 3! 5! x x x ≤ −
+ . 1-teorema isbotlandi. 2-teorema. n = k (k - juft) da (9.3) tengsizlikning bajarilishi berilgan:: 68
3 2 1
3 2 1 ... sin
... , 0
3! (2 1)! 3! (2 1)! 2 k k x x x x x x x x k k π − + − + − ≤ ≤ −
+ + ≤ ≤
− + . (9.3.6) n=k+2 da ushbu tengsizlikninig bajarilishini isbotlash lozim. 3 2 3 3 2 5 ...
sin ...
, 0 3! (2 3)! 3! (2 5)! 2
k x x x x x x x x k k π + + − + − ≤ ≤ −
+ + ≤ ≤
+ + . (9.3.7) Isbotlash. 1-teoremani isbotiga o’xshash to’rt marotaba X ni T ga almashtirish va T bo’yicha 0 dan 0, 2
π ⎛ ⎞ ∈⎜ ⎟ ⎝ ⎠ gacha mos tengsizlikni integrallash protsedurasini bajarish kerak. (9.3.6) tengsizlikning o’ng qismida x ni t ga almashtirish va T bo’yicha 0 dan 0, 2
π ⎛ ⎞ ∈⎜ ⎟ ⎝ ⎠ gacha ushbu tengsizlikni integrallab quyidagini hosil qilamiz: ) 2 2 ( ... ! 4 2 ! ) 1 2 ( ... ! 3 sin cos 1 2 2 4 2 0 1 2 3 0 + + + − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + − ≤ = − + + ∫ ∫
x x x dt k t t t dt t x k x k x . ) 2 2 ( ... ! 4 2 1 cos 2 2 4 2 + − − + − ≥ ⇒ + k x x x x k . Ikkinchi marotaba X ni T ga almashtirish va T bo’yicha 0 dan 0, 2 x π ⎛ ⎞ ∈⎜ ⎟ ⎝ ⎠ gacha integtallab, quyidagiga ega bo’lamiz: 2 2 2 3 2 3 0 0 sin cos
1 ...
... 2 (2 2)! 3! (2 3)! x x k k t t x t x tdt dt x k k + + ⎛ ⎞ = ≥ − + − = − + −
⎜ ⎟ + + ⎝ ⎠ ∫ ∫ . Bu quyidagiga teng kuchli
3 2
sin ...
3! (2 3)! k x t x x k + ≥ − + − + , 0, 2
π ⎛
∈⎜ ⎟ ⎝ ⎠ . (9.3.8) (9.3.6) tengsizlikning chap qismi isbotlandi. (9.3.8) tengsizlikni uchinchi marotaba X ni T ga almashtirish va T bo’yicha ⎟ ⎠
⎜ ⎝ ⎛ ∈ 2 , 0 π
gacha integrallab, quyidagiga ega bo’lamiz: 69
3 2 3 0 0 1 cos sin ...
3! (2 3)! x x k t t x tdt t dt k + ⎛ ⎞ − = ≥ − + −
= ⎜ ⎟ + ⎝ ⎠ ∫ ∫ 2 4 2 4 ... 2 4! (2 4)!
k x x x k + − + − + . Bundan 2 4 2 4 cos 1 ...
2 4! (2 4)! k x x x x k + ≤ − + − +
+ . To’rtinchi marotaba ushbu tengsizlikni X ni T ga almashtirish va T bo’yicha 0, 2 x π ⎛ ⎞ ∈⎜ ⎟ ⎝ ⎠ gacha integrallab, quyidagiga ega bo’lamiz: 2 2 4 3 2 5 0 0 sin cos
1 ...
... 2 (2 4)! 3! (2 5)! x t k k t t x x x tdt dt x k k + + ⎛ ⎞ = ≤ − + + = − + +
⎜ ⎟ + + ⎝ ⎠ ∫ ∫ . Ya’ni quyidagi tengsizlik bajariladi: 3 2 5 sin
... 3! (2 5)! k x x x x k + ≤ − + + + . (9.3.6) tengsizlikning o’ng qismi, ushbu tengsizlik va (9.3.8) tengsizlikdan (9.3.7) tengsizlik hosil qilinadi. 2-teorema isbotlandi. 9.4-masala. Har bir natural n > 1 uchun 2 2 1 n + soni 7 raqami bilan tugallanadi. 1-teorema. n=2 da 7 raqami bilan tugallanuvchi 2 2 2 1 17
+ = hosil bo’ladi. 2-teorema. 2 2 1 k + sonning 7 raqami bilan tugallanishi berilgan. 1 2 2 1 k + + sonning 7 raqami bilan tugallanishini isbotlash losim. Isbotlash. 2 2 1 k + son 7 raqami bilan tugallanganligi sabab, 2 2
soni 6 raqami bilan tugaydi. Agar son 6 raqami bilan tugallansa, u holda unung kvadrati ham 6 raqami bilan tugaydi. n=k+1 da quyidagini hosil qilamiz:
k k k 2 2 2 2 1 2 2 2 2 2 ⋅ = = ⋅ + va 6 raqami bilan tugallanadi. 2-teorema isbotlandi. 9.5-masala. Ixtiyoriy n naturl son uchun ( ) ( ) 1 1 10 10 ... 1 10 5 1 n n n − + + + + ⋅
+ + soni to’liq kvadrat ekanligini isbotlang. 70
Isbotlash. Ushbu masalani yechishda matematik induksiya metodidan foydalanmasdan bajarish mumkin. 1 10
... 1 n n − + + + yig’indi q=10 maxrajli n+1 hadlardan iborat geometric progressiyani anglatadi. U holda ( ) ( ) ( ) 1 1 1 1 10 1 10 10 ... 1 10 5 1 10 5 1 10 1 n n n n n + − + + ⎛ ⎞ − + + + ⋅ + + =
⋅ + + =
⎜ ⎟ − ⎝ ⎠
( ) ( ) 2 2 2 1 1 1 1 1 2 1 10 10 5 10
5 9 10 2 10 2 2 10 2 9 9 3 n n n n n n + + + + + + − + ⋅ − + + ⋅
⋅ + ⎛ ⎞ + = = = ⎜ ⎟ ⎝ ⎠
ixtiyoriy n natural son uchun. 2-teorema isbotlandi. Eslatma. Yuqorida ko’rilgan masalalardan matematik induksiya metodi yordamida juda katta sinfdagi turli masalalarni yechish ko’rsatiladi. Ushbu metodni qo’llashga doir ko’p masalalarni ko’rsatish mumkin. Masalan quyidagi tengsizlikni isbotlaylik:
1 1 1 1 1 ...
, : 2 1 2 2 1 2 2
n n n n + + + + < ∀ ∈
≥ + + − .
tengsizlikning bajarilishini isbotlash lozim. 1 1 1 1 ... 1 2 2 1 2 n S n n n n = + + + + + + − ni belgilaymiz. U holda quyidagi hosil qilinadi: 1 1 1 1 1 1 1 1 1 1 ... 1 2 2 1 2 2 1 2 2 2 2
1 2 2
k S S k k k k k k k k + = = + + + + + + < + + + + − + + + + . Hosil bo’lgan tengsizlikning chap qismi 0,5 dan kichik bo’lganligi sababli natijaga erishib bo’lmaydi. Matematik induksiya metodini qo’llash qiyinchiliklarga olib keldi. Bu tengsizlikni boshqa usul bilan isbotlanadi. Ixtiyoriy 2
≥ quyidagi navbatdagi n -1 tengsizlik va tenglik bajariladi 71
1 1 1 1 1 1 1 1 , , ... , 1 2 2 2 2 1 2 2 2 n n n n n n n n > > > = + + − . Ushbu tengsizlik va tengliklarni qo’shib talab etilayatgan quyidagi tengsizlikni hosil qilamiz 1 1
1 1 1 ... 1 2 2 1 2
2 2
S n n n n n n = + + + + > ⋅ = + + − .
72
Mustaqil yechish uchun masalalar. 1. Arifmetik progressiyaning n- hadi quyidagi formula bilan hisoblanishini isbotlang
– birinchi had, d – progressiya ayirmasi. 2. Geometrik progressiyaning n-hadi quyidagi formula bilan hisoblanishini isbotlang.
a n = a 1 q n – 1 , bu yerda a 1 – birinchi had ,
Quyidagilarni isbotlang: 1.
3 3 3 ( 1) ( 2) , ,
n n n N + +
+ + ∀ ∈
9 ga bo’linadi. 2.
3 2 , n n n n N − ≥ ∀ ∈ . 3.
2 2 1 1 ,
n n n N n n ⎛ ⎞ + + ≥ ∀ ∈ ⎜ ⎟ ⎝ ⎠ . 4. 2 2 2 2 1 2 1 ( 1) 1 2 3 4 ... ( 1)
( 1) , 2 n n n n n n N − − + − + − + + −
= − ∀ ∈
. 5.
( 1) (
2) 1 2 2 3 3 4 ... ( 1)
3 n n n n n n N ⋅ + ⋅ +
⋅ + ⋅ + ⋅ + + ⋅ + = ∀ ∈
. 6.
1 1 1 1 ...
, 4 5 5 6 6 7 ( 3) (
4) 4 (
4) n n N n n n + + + + = ∀ ∈ ⋅ ⋅ ⋅ + ⋅ + ⋅ +
. 7.
1 1 1 1 ...
, 1 4 4 7 7 10 (3 2) (3
1) (3 1) n n N n n n + + + + = ∀ ∈ ⋅ ⋅ ⋅ − ⋅ + + . 8.
N n n n n n ∈ ∀ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + > ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + + , 1 1 1 1 1 2 1 . 9. 1 1 1 1 1 , 1 n n n N n n + ⎛ ⎞ ⎛ ⎞ − < − ∀ ∈
⎜ ⎟ ⎜ ⎟ + ⎝ ⎠ ⎝ ⎠ . 10.
1 1! 2 2! 3 3! ... ! (
1)! 1, n n n n N ⋅ + ⋅ + ⋅ + + ⋅ = + − ∀ ∈ .
2 2 2 4 (2 1)(2
1) 2 6 ... (4 2) , 3 n n n n n N − + + + +
− = ∀ ∈ . 73
12. 7 7 7 7 1 ... 1 , 1 8 8 15 15 22 (7 6) (7 1) (7 1) n N n n n + + + + = −
∀ ∈ ⋅ ⋅ ⋅ − ⋅
+ + . 13. 1 1 1 1 1 1 ...
, 4 8 8 12 12 16 4 (4 4) 16 16 ( 1) n N n n n + + + + = − ∀ ∈ ⋅ ⋅ ⋅ ⋅ + ⋅ + . 14. 5! 6! (5 )! (5 1)! ...
, 0! 1!
! 6 !
n n n N n n + + + + + +
= ∀ ∈
. 15.
2 ( 1) (1 ) 1 , : 3, 0 2
n n a na a n N n a − + > + + ∀ ∈ ≥ > . 16. 2 : , ! ) 1 2 ( 1 2 ≥ ∈ ∀
− −
N n n n n .
17. N n n n n ∈ ∀ − = − + + + + ), 1 2 ( ) 1 2 ( ... 5 3 1 2 2 3 3 3 3 . 18. 3 ln 1 1 1 3 ln 3 ln 1 ...
27 ln 9 ln 1 9 ln 3 ln 1 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⋅ + + ⋅ + ⋅ −
n n , : 2 n N n ∀ ∈
≥ . 19. Ixtiyoriy n natural son uchun 1 2 2 12 11 − + +
n ifodaning 133 ga karrali ekanligini isbotlang. 20.
2 2 1 1 6 3 3 , n n n n N − + − + + ∀ ∈ sonning 11 ga bo’linishini isbotlang. 21.
∈ ∀ + + , 2 10 1 sonning 3 ga bo’linishini isbotlang. 22. 7 tiyindan katta bo’lgan ixtiyori pullar yig’indisini 3 va 5 tiyinliklar bilan qaytimsiz to’lash mumkinligini isbotlang. 23. Faraz qilaylik ketma-ketlik quyidagicha berilgan bo’lsin:
1 2 1 2; 3; 3 2 , : 3
n n x x x x x n N n − = = = − ∀ ∈ ≥ , 1 2 1, n n x n N − = + ∀ ∈
formulaning to’g’riligini isbotlang. 24. Tengsizlikni isbotlang: a).
,...,
1 ], , 0 [ , , sin
sin 1 1 = ∈ ∈ ∀ ≤ ∑ ∑ = = π . b). ) , ( , , sin sin ∞ + −∞ ∈ ∈ ∀ ≤
N n x x n .
74
c). N n x x n n ∈ ∀ ≤ + , 1 cos
sin 2 2 . 25. Tengsizlikni isbotlang: 1 21
5 ... . 5 , 2 n n ildiz x n N + = + + +
< ∀ ∈
. 26. Tenglikni isbotlang: 2 ( 1) sin
sin ... sin
2sin sin
, 3 3 3 6 3 n n n n N π π π π π + + + + = ⋅ ∀ ∈ . 27.
2 1 sin 1 2 cos cos 2 ... cos
, 2 , 1 2 2sin 2 n x x x nx n k n N x π + + + + + = ≠ ∀ ∈ .
2 1
1 1 , , , ... ,
, ... 3 15 35
4 1
− ketma-ketlik berilgan. Dastlabki n hadi yig’indisini toping. 28. Har qanday n natural son uchun
− 5 ifoda 5 ga karrali ekanligini isbotlang.
− ning k ga karrali mulohazasining bajarilishini tekshiring. 29. Ixtiyoriy n ∈
a). 1
2 2 3 3 6 − + − + + n n n sonning 11 ga bo’linishini. b). 1 3 2 3 3 2 5 − − + ⋅ n n sonning 19 ga bo’linishini. 30. Ixtiyoriy
∈
a). 1 3 1 1 2 1 ...
2 1 1 2 +
− +
+ < n n n . b). Agar ) 1 ( 0 n i x i ≤ ≤ > i
2 1 ... 2 1 ≤ + + + n x x x bo’lsa, u holda
2 1 ) 1 ( ... ) 1 ( ) 1 ( 2 1 ≥ − ⋅ ⋅ − ⋅ − n x x x .
75
c). Agar ) 1 ( 0
i x i ≤ ≤ > va
10 ...
2 1 = + + + n x x x , bo’lsa, u holda
⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≤ ⋅ ⋅ ⋅ 10 ... 2 1 . d). 2 )
( ! ) 2 ( 1 4 n n n n > + . 31. Ixtiyoriy n ∈
a) 0 , ... ...
2 1 2 1 > + + + ≤ ⋅ ⋅ ⋅ i n n x n x x x x x x ; b) ( )
i x n x x x x x x i n n ...,
, 2 , 1 , 0 , 1 ... 1 1 ... 2 2 1 2 1 = > ≥ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + + ⋅ + + + ; c) 0 ,
2 2 2 1 2 1 2 1 > ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ≥ + x x x x x x n n n ;
d) 1 1 1 1 ... 2 4 4 2 + ≥ + + + + + + − − − − n x x x x x x n n n n n n .
∈
lg1 lg 2 ... lg lg( 1)
n n + + + + > tengsizlikni isbotlang. 76
Adabiyotlar
1. Брадис В.М., Минковский В. Л., Харчева А. К. Ошибки в математиче- ских рассуждениях. — М.: Учпедгиз, 1959. 2. Уфановский В. А. Математический аквариум. — Ижевск: НИЦ «Регу- лярная и хаотическая механика», 2000. 3. Соловьев Ю. П. Задачи по алгебре и теории чисел для математических школ. Ч. 1 - 3. — М.: школа им. А. Н. Колмогорова, 1998. 4. Головина Л. И., Яглом И. М. Индукция в геометрии. Серия «популярные лекции по математике» — Вып. 21.— М.: Наука, 1961. 5. Соминский И. С. Метод математической индукции. Серия «популярные лекции по математике» — Вып. 3. — М.: Наука, 1974. 6. Успенский В. А. Треугольник Паскаля. Серия «Популярные лекции по математике» — Вып. 43. — М.: Наука, 1979. 7. Аврамов А. Арифметические прогрессии в треугольнике Паскаля // Квант, №11. 1980. 8. Бендукидзе А. Треугольник Паскаля // Квант, № 10. 1982. 9. Бендукидзе А., Сулаквелидзе А. Вычисление сумм // Квант, № 9. 1970. 10. Кузьмин Е., Ширшов А. О числе е // Квант, № 8. 1979. 77
Mundarija
1-§. Matematik induksiya metodi haqida………………………….. 3 2-§. Tengliklarni isbotlash………………………………………….. 7 3-§. Xaritani bo’yash......................................................................... 20 4-§. Tengsizliklarni isbotlash............................................................ 22 5-§. Gipoteza va uning isbotlanishi……………………………….. 40
6-§. Paskal uchburchagi…………………………………………… 51
7-§. Nyuton binomi formulasi…………………………………….. 53
8-§. Natural sonlarning bo’linishini isbotlash……………………... 56
9-§. Turli masalalar........................................................................... 58 Mustaqil yechish uchun masalalar………………………………… 67 Adabiyotlar…………………………………………………………. 71 Document Outline
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