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![2) Cnk=Cnn−k - bu tenglik va 1) dan binom koeffitsentlari boshidan o`rtasigacha
bo`lgan elementlari, o`rtasidan keyingi elementlari bilan bir xilligini bildiradi va
demak ularning yig`indilari ham teng.
Shularga asoslanib n ni ikki holda qaraymiz.
1) n = 2k + 1 , ya`ni toq bo`lsin.
C
2 k + 10
+ 2 C
2 k + 11
+ … +
( k + 1 ) C
2 k + 1k
+ ( k + 2 ) C
2 k + 1k + 1
+ … + ¿
+
( 2 k + 1 ) C
2 k + 12 k
+ + ( 2 k + 2 ) C
2 k + 12 k + 1
= { C
2 k + 10
+ ( 2 k + 2 ) C
2 k + 12 k + 1 }
+ ¿
+{2Ck+11 +(2k+1)C2k+1 2k }+… +{(k+1)C2k+1 k +(k+2)C2k+1 k+1}=¿
(C2k+1 0 =C2k+1 2k+1,C2k+1 1 =C2k+1 2k ,… ,C2k+1 k =C2k+1 k+1larga ko 'ra )
¿
( 2 k + 3 ) C
2 k + 10
+ ( 2 k + 3 ) C
2 k + 11
+ … + ( 2 k + 3 ) C
2 k + 1k
= ¿
¿(2k+3)[C2k+1 0 +C2k+1 1 +… +C2k+1 k ]=(2k+3)22k+1
2 =(2k+3)22k=¿
¿(n+2)2n−1
chunki, 2k + 1 = n dan
( 2 k + 3 ) 2 2 k
= ( n + 2 ) 2 n − 1
va
C
2 k + 10
+ C
2 k + 11
+ … + C
2 k + 1k
+ C
2 k + 1k + 1
+ … + C
2 k + 12 k + 1
= 2 2 k + 1
C
2 k + 10
+ C
2 k + 11
+ … + C
2 k + 1k
= C
2 k + 1k
+ … + C
2 k + 12 k + 1
= 2 2 k + 1
2 = 2 k
dan foydalandik.
2) n = 2k bo`lsin
1C2k0+2C2k1+… +(k+1)C2kk+(k+2)C2kk+1+… +(2k+1)C2k2k=¿
¿
{ 1 ∙ C
2 k0
+ + ( 2 k + 1 ) C
2 k2 k }
+ { 2 C
2 k1
+ 2 k C
2 k2 k − 1 }
+ … + ¿
+
{ k C
2 kk − 1
+ ( k + 2 ) C
2 kk + 1 }
+ ( k + 1 ) C
2 kk
= ¿
¿(2k+2)C2k0+(2k+2)C2k1+… +(2k+2)C2kk−1+(k+1)C2kk=¿
¿ 2
( k + 1 )[ C
2 k0
+ + C
2 k1
+ … C
2 kk − 1 ]
+ ( k + 1 ) C
2 kk
(*)
C2k0+C2k1+… C2kk−1+C2kk+C2kk+1+… +C2k2k=22k
va C
nm
= C
nn − m
dan binom koeffitsentlari o`rtasiga nisbattan simmetrik
joylashganidan
C
2 k0
+ C
2 k1
+ … C
2 kk − 1
+ C
2 kk
2 = 2 2 k
2 = 2 2 k − 1
Shularga asosan (*) tenglikda
17](https://docx.uz/documents/e86e7518-4075-4127-a2e2-233d573c3adb/page_17.png?v=1)
![2(k+1)[C2k0+C2k1+… C2kk−1]+(k+1)C2kk= ¿¿ 2
( k + 1 )[ C
2 k0
+ C
2 k1
+ … + + C
2 kk − 1
+ C
2 kk
2 ] = ¿
¿
( 2 k + 2 ) 2 2 k − 1
= ( n + 2 ) 2 n − 1
.
Shunday qilib formula bitta usulda isbotlandi.
II – usul.
Cn0+2Cn1+… +(n+1)Cnn=¿
¿ C
n0
+ C
n1
+ … + C
nn
+ C
n1
+ 2 C
n2
+ … + +
( n − 1 ) C
nn − 1
+ nC
nn
= ¿
¿2n+n+2n(n−1)
1∙2 +3n(n−1)(n−2)
1∙2∙3 +… +n−1
1 +nC nn= ¿
¿2n+n(1+n−1
1 +(n−1)(n−2)
1∙2 +… +(n−1)(n− 2)… 2∙1
(n− 2)… 2∙1 +1)= ¿
¿2n+n(Cn−10 +Cn−11 +… +Cn−1n−2+Cn−1n−1)=¿
¿ 2 n
+ n ∙ 2 n − 1
=
( n + 2 ) 2 n − 1
.
III – usul. Matematik induksiya usulidan foydalanamiz.
C
n0
+ 2 C
n1
+ … +
( n + 1 ) C
nn
= ( n + 2 ) 2 n − 1
n = 1 da
C10+2C11=1+2= 3
(
n + 2 ) 2 n − 1
= ( 1 + 2 ) ∙ 2 0
= 3
n = k da
Ck0+2Ck1+… +(k+1)Ckk=(k+2)∙2k−1
tenglik o`rinli bo`lsin. n = k + 1 da tenglikni ko`rsatamiz.
C
k + 10
+ 2 C
k + 11
+ 3 C
k + 12
+ … + ( k + 2 ) C
k + 1k + 1
ni quyidagi formula yordamida shakl almashtiramiz.
Cnm+Cnm+1=Cn+1m+1
ga ko`ra
C
k + 10
+ 2 C
k + 11
+ 3 C
k + 12
+ … +
( k + 2 ) C
k + 1k + 1
= C
k + 10
+ 2 ( C
k0
+ C
k1 )
+ ¿
+3(Ck1+Ck2)+… +(k+1)(Ckk−1+Ckk)+k+2= ¿
¿Ck+10 +2Ck1+3Ck2+… +(k+1)Ckk+2Ck0+3Ck1+… +¿
+(k+1)Ckk−1+k+2= (k+2)∙2k−1+¿
+ C
k0
+ 2 C
k1
+ … + k C
kk − 1
+
( k + 1 ) C
kk
− ( k + 1 ) C
kk
+ k + 2 + ¿
18](https://docx.uz/documents/e86e7518-4075-4127-a2e2-233d573c3adb/page_18.png?v=1)
![100 ga bo`linadi va bu ayirma yana berilgan x1,x2,… ,x100 larning qandaydir
yig`indisidan iborat bo`ladi.
Demak, masala hal bo`ldi.
10 – misol.
1p+2p+… +np soni p – toq bo`lganda, 1 + 2 + … + n
soniga
bo`linishini isbotlang.
Isbot. Shu ko`rinishda misolni yechish qiyin. Berilgan
1+2+… +n ni n ( n + 1 )
2
ko`rinishda ifodalaymiz. n va ( n + 1 )
sonlari o`zaro tub bo`lgani uchun
1p+2p+… +np ni alohida n ga va ( n + 1 )
ga bo`linishini ko`rsatsak yetarlidir.
1) Agar n – toq bo`lsa,
1 p
+ 2 p
+ … + n p
ni quyidagicha gruppalaymiz
1 p
+ ( n − 1 ) p
+ 2 p
+ ( n − 2 ) p
+ … + k p
+ ( n − k ) p
+ … + n p
Har bir k da
kp+(n− k)p= n(kp−1−… +(n− k)p−1) ga ko`ra n ga bo`linadi.
Endi n + 1
2 ga bo`linishini ko`rsatamiz, n = 2k – 1
1 p
+ 2 p
+ … + k p
+ ( k + 1 ) p
+ … + ( 2 k − 1 ) p
= ¿
[
1 p
+ ( 2 k − 1 ) p ]
+ + [ 2 p
+ ( 2 k − 2 ) p ]
+ … + ¿
+[(k−1)p+(k+1)p]+kp,
1 p
+ ( 2 k − 1 ) p
= 2 k ∙ A
1
va hokazo
(k−1)p+(k+1)p=2k∙Ak
Ak= kp−1−… +(k+1)p−1
larga asosan har bir juftlik n + 1
2 = 2 k
2 = k
ga bo`linadi va oxirgi
kp ham n + 1
2 = 2
ga
bo`linadi.
2) n – juft son bo`lsa, ya`ni n = 2k bo`lsa
1p+2p+… +kp+(k+1)p+(k+2)p+… +¿
+ ( 2 k − 1 ) p
+ ( 2 k ) p
=
[ 1 p
+ ( 2 k − 1 ) p ]
+ ¿
+[2p+(2k−2)p]+… +(2k)p.
25](https://docx.uz/documents/e86e7518-4075-4127-a2e2-233d573c3adb/page_25.png?v=1)
![dan har bir qo`shiluvchining 2k ga bo`linishi kelib chiqadi, ya`ni n ga bo`linishi
kelib chiqadi. Endi ( n + 1 ) ga yoki n+1
2 ga bo`linishini ko`rsatamiz.
1p+2p+… +kp+(k+1)p+(k+2)p+… +¿
+(2k− 1)p+(2k)p=[1p+(2k)p]+¿
+[2p+(2k−1)p]+… +[kp+(k+1)p],
har bir qo`shiluvchi 2k + 1 ga bo`linadi, ya`ni n + 1 ga bo`linadi. Shunday qilib
berilgan yig`indi o`zaro tub sonlarning har biriga bo`linsa, ularning ko`paytmasiga
ham bo`linishidan masalani yechgan bo`lamiz.
11 – misol.
( n + 1 )( n + 2 ) ∙ … ∙ 2 n
sonining 1 ∙ 3 ∙ 5 ∙ … ∙ ( 2 n − 1 )
soniga bo`linishini
isbotlang.
Isbot. Teskarisidan faraz qilamiz, ya`ni
Sn=(n+1)∙… ∙2n
va Cn=1∙3∙… ∙(2n−1)
sonlari karrali emas, ya`ni
Sn≠ pCn , p – butun son yoki
Sn
Cn – butunmas ekan
deylik.
Sn
Cn ni sur`at va maxrajini
2∙4∙… ∙2n ga ko`paytiramiz:
Sn
Cn
= 2∙4∙… ∙2n∙Sn
2∙4∙… ∙2n∙1∙3∙… ∙(2n−1)= 2n∙n!∙Sn
(2n)! = 2n∙(2n)!
(2n)! = 2n
bu faramizga ziddir, demak
Sn
Cn - butundir. Masala yechildi.
Natija hosil bo`ladi.
Natija.
(n+1)(n+2)∙… ∙(n+n)=2n∙1∙3∙… ∙(2n−1) formula o`rinli.
Lemma: x>0 va y>0 sonlari uchun 1
x + 1
y ≥ 4
x + y tengsizlik o‘rinli bo‘ladi.
Isbot: Haqiqatdan ham berilgan tengsizlikni quyidagicha shakl almashtirsak:
1
x + 1
y ≥ 4
x + y => x + y
xy ≥ 4
x + y =>
(x+y)2≥4xy =>
x 2
+ 2 xy + y 2
− 4 xy ≥ 0 => =>
( x − y ) 2
≥ 0
da’vo isbot bo‘ldi. Bu tengsizlikni umumlashtirsak, ya’ni a,b,x,y>0 sonlari uchun
26](https://docx.uz/documents/e86e7518-4075-4127-a2e2-233d573c3adb/page_26.png?v=1)
![a2
x+b2
y ≥(a+b)2
x+y tengsizlik o‘rinli bo‘ladi. Haqiqatdan ham, (
a2x+b2y¿(x+y)≥≥xy (a+b)2
=>
a 2
x 2
+ a 2
xy + b 2
xy + b 2
y 2
≥ a 2
xy + 2 abxy + b 2
xy =>
a 2
x 2
− 2 abxy + b 2
y 2
≥ 0 =>
( ax − by ) 2
≥ 0 bo ' ladi .
4 §. Sonning butun qismi bilan bog`liq masalalar.
x sonning butun qismi deb, x dan katta bo`lmagan va x ga eng yaqin
butun songa aytiladi.
Ta`rifga ko`ra agar x ning butun qismini
[x] bilan belgilasak, [x]≤xva
[x]+1>x
.
Sonning butun qismining quyidagi xossalarini keltiramiz:
10.
[x]= n - butun son
2 0
. x = n + α
,0≤α<1.
30.
Agar [x]=[y] bo`lsa, x = n + α, y = n + β, α, β ϵ
[0,1¿
- 1 < x – y = α – β < 1
40.
[x+y]= x bo`lsa, x – butun va 0 ≤ y < 1
.
5 0
. n – butun bo`lsa,
[ n + x ] = n + [ x ]
1 – misol.
[
8x+19
7 ]= 16 (x+1)
11 tenglamani yeching.
Yechish. Modul xossasidan
0 ≤
[ 8 x + 19
7 ] − 16 ( x + 1 )
11 < 1
0 ≤
88 x + 209 – 112 x – 112 < 77
97
≤ - 24 x < - 20
5
6 < x
≤ 4 1
24
Bu oraliqdan qaysi sonni olsak 16 ( x + 1 )
11 butun chiqishini aniqlash qiyin.
27](https://docx.uz/documents/e86e7518-4075-4127-a2e2-233d573c3adb/page_27.png?v=1)
![Shuning uchun 16 (x+1)
11 =t - butun son deb, x = 11 t − 16
16 ni topamiz va berilgan
tenglamani t ga nisbattan yozamiz:
[
11 t+22
14 ]=t , t ni butunligidan foydalansak va
0≤11 t+22
14 − t<1
, 0≤22 −3t<14 , −22 ≤−3t← 8 , 22
3<t≤71
3
Bu yerdan t = 3, 4, 5, 6, 7 ni olamiz va bu qiymatlarni ketma – ket
x = 11 t − 16
16 ga qo`ysak
x=1 1
16 ,13
4,2 7
16 ,31
8,313
16 yechimlarni olamiz.
Sonning butun qismi yana quyidagi muhim xossalarga ega.
1 – Teorema. Agar n – manfiymas butun son bo`lsa,
[nx ]≥n[x] .
Isbot. x = m + α , m – butun, 0 ≤ α < 1
bo`lsin, nx = nm + nα
[nx ]= nm +[nα ]≥nm =[nx ]
2 – Teorema. Har qanday N, q
ϵ N uchun [
N
q ]∙q≤N .
Isbot. Isbot ta`rifdan kelib chiqadi, ya`ni
[
N
q ]∙q≤N ⟹ [
N
q ]≤ N
q
,
ya`ni sonning butun qismi shu sonning butun qismidan oshmaydi.
3 – Teorema. Har qanday N, q
ϵ N uchun ([
N
q ]+1)∙q>N .
Buning ham isboti ta`rifdan kelib chiqadi, ya`ni
([
N
q ] + 1 ) ∙ q > N
⟹ [ N
q ] + 1 > N
q
4 – Teorema.
1,2,3,… ,n – natural sonlar ketma – ketligida qϵN soniga
bo`linadiganlari soni
[
n
q] ga teng.
Isbot. Agar q > n bo`lsa,
[
n
q]= 0 , ya`ni 1, 2, … , n – ketma – ketlik hadlari
q ga bo`linmaydi. q = n bo`lsa, bu ketma – ketlik ichida faqat n, q ga bo`linadi.
Bu holda ham teorema o`rinli.
Endi q < n holni qaraymiz.
28](https://docx.uz/documents/e86e7518-4075-4127-a2e2-233d573c3adb/page_28.png?v=1)
![1, 2, … , n ketma – ketlik ichida q ga bo`linadiganlar m ta bo`lsin, ya`ni 1∙q,2∙q,… ,m∙q
. m=[
n
q] ekanligini ko`rsatamiz. Teskarisidan faraz qilaylik, ya`ni
m≠[
n
q]
. Bu holda 2 hol bo`lishi mumkin.
m<[
n
q] bo`lsa, [ n
q ] ∙ q ≤ n
tengsizlikdan va [ n
q ] ∙ q
ning q ga bo`linishidan q
ga bo`linadiganlar soni m dan ko`p ekanligi kelib chiqadi. Bu bo`linuvchilar m
ta degan tasdiqqa zid.
m >
[ n
q ] bo`lsin. Bu holda m ≥ [ n
q ] + 1
degani. Oxirgi tengsizlikni ikkala
qismini q ga ko`paytirsak mq ≥
([ n
q ] + 1 ) q > n
ga ega bo`lamiz. Bu mq > n , ya`ni 1,
2, … , n qatorda yotmasligini bildiradi. Bu zidliklardan m =
[ n
q ] ekanligi kelib
chiqadi.
Teorema isbot bo`ldi.
Yuqorida keltirilgan 4 – teoremada q – natural son ixtiyoriy tanlangani
uchun, quyidagi natijaga kelamiz.
Natija. 1 , 2 , … , n
– natural sonlar ketma – ketligida q ga karralilar
[ n
q ] ta,
q 2
ga bo`linadiganlar
[ n
q 2 ] ta,
q 3
karralilar [ n
q 3 ] ta va hokazo.
5 – Teorema. n! sonida q – tub soni nechanchi darajada qatnashishi
quyidagi formula yordamida topiladi:
[
n
q ] + [ n
q 2 ] + [ n
q 3 ] + … + [ n
q m ] ,
bunda
q m
≤ n , q m + 1
> n .
Isboti natijadan kelib chiqadi.
29](https://docx.uz/documents/e86e7518-4075-4127-a2e2-233d573c3adb/page_29.png?v=1)
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