Paskal uchburchagi. nyuton binomi. binomial koeffitsiyent
![Agar n toq son bo‘lsa, m= n−1
2 butun son bo‘lib,
n− m
m +1=
n− n− 1
2
n− 1
2 +1
= 2n− n+1
n− 1+2 = n+1
n+1= 1
munosabat o‘rinlidir. Demak,
Cnm+1= n−m
m+1Cnm
formuladan
m= n−1
2 bo‘lganda Cn
n−1
2 +1
=Cn
n−1
2 tenglik kelib chiqadi.
Binomial koeffitsientlarning 5- xossasi Paskal uchburchagining yuqorida
keltirilgan xossalari tasdig‘i bo‘lib, unga ko‘ra binomial koeffitsientlar oldin
Cn0=1
dan
Cn
[n2] gacha 11
o‘sadi, keyin esa Cnn=1 gacha kamayadi hamda n toq bo‘lganda
binomial koeffitsientlar qatorining o‘rtasidagi ikkita hadi tengdir va
n juft
bo‘lganda uning o‘rtadasigi hadi eng katta va yagonadir.
Quyidagi 6–8- xossalar o‘rinlidir.
6- x o s s a .
Cnn+Cn+1n +...+Cn+kn =Cn+k+1 n+1 .
7- x o s s a .
(C n0)2+(Cn1)2+...+(Cnn)2= C2nn .
8- x o s s a .
Cn0Cmk+Cn1Cmk−1+...+CnkCm0=Cn+mk .
Oxirgi tenglik Koshi 12
ayniyati deb aytiladi.
Endi bu uchta xossalarni isbotlaymiz. Dastlab 6- xossaning isbotini
keltiramiz. Birinchidan,
s=(1+x)n+(1+x)n+1+...+(1+x)n+k
ko‘phad uchun Nyuton binomi formulasini qo‘llab, quyidagi tenglikni hosil qilamiz:
s= ∑m=0
n
Cnmxm+∑m=0
n+1
Cn+1m xm+...+∑m=0
n+k
Cn+km xm
.
Bu yerdan,
s ko‘phaddagi xn ifoda ning koeffitsienti
Cnn+Cn+1n +...+Cn+kn
yig‘indiga tengligini aniqlash mumkin.
11
[a] yozuv a son ning butun qismini anglatadi.
12
Koshi (Cauchy Ogyusten Lui, 1789-1857) –fransuz matematigi.](https://docx.uz/documents/7a24074e-92fe-4a17-bfab-173d43d69571/page_8.png?v=1)
![Oxirgi tenglikning ikkala tomonini (−1 )ga ko‘paytirsak, uk+12 −u(k+1)−1u(k+1)+1=(−1)(k+1)+1
tenglik hosil bo‘ladi.
Matematik induksiya usulini qo‘llab
u1,u2,... Fibonachchi sonlarining
quyidagi 7–10- xossalarni ham isbotlash mumkin:
7- x o s s a .
u1u2+u2u3+u3u4+...+u2n−1u2n=u2n2 .
8- x o s s a .
u1u2+u2u3+u3u4+...+u2nu2n+1= u2n+1 2 −1 .
9- x o s s a .
nu 1+(n−1)u2+(n−2)u3+...+2un−1+un=
= un+4−(n+3)
.
10- x o s s a .
u1+2u2+3u3+...+nu n= nu n+2− un+3+2 .
Endi Fibonachchi sonlarining binomial koeffitsientlar (Paskal uchburchagi)
bilan bog‘lanishini ifodalovchi xossani o‘rganamiz.
11- x o s s a . Fibonachchi soni
un ( n∈N ) uchun un= ∑k=0
[n−12]
Cn−k−1 k tenglik
o‘rinlidir .
Bu xossani isbotlash uchun
un ( n=1,2 ,... ) sonlardan tuzilgan u1,u2,...,un,...
ketma-ketlikning Fibonachchi qatori bo‘lishini ko‘rsatish kifoya. Buning uchun esa
u1= ∑k=0
[1−12]
C1−k−1 k =∑k=0
0
C−kk=C00=1
,
u2= ∑k=0
[2−12]
C2−k−1 k = ∑k=0
0
C1−kk =C10=1
ekanligini ta’kidlab,
u1,u2,...,un,... ketma-ketlik uchun un+1=un+un−1 ( n≥2 ) rekurrent
tenglikning bajarilishini ko‘rsatamiz.
Agar
n juft son ( n=2s , s∈N ) bo‘lsa, u holda
un+1=∑k=0
[n2]
Cn−kk =∑k=0
s
Cn−kk
,
un= ∑k=0
[n−12]
Cn−k−1 k =∑k=0
s−1
Cn−k−1 k
,
un−1= ∑k=0
[n−22]
Cn−k−2 k =∑k=0
s−1
Cn−k−2 k
tengliklar o‘rinli bo‘ladi. Bu tengliklardan foydalanib,](https://docx.uz/documents/7a24074e-92fe-4a17-bfab-173d43d69571/page_14.png?v=1)
![un+un−1=∑
k=0
s−1
Cn−k−1 k +∑
k=0
s−1
Cn−k−2 k =1+∑
k=1
s−1
Cn−k−1 k +∑
p=1
s
Cn−p−1 p−1 =
=1+∑k=1
s−1
Cn−k−1 k +∑p=1
s−1
Cn−p−1 p−1 +Cn−s−1 s−1 =
=1+∑k=1
s−1
(Cn−k−1 k +Cn−k−1 k−1 )+Cn−s−1 s−1munosobatlarni hosil qilamiz . Binomial koeffitsientlarning
Cn−k−1 k +Cn−k−1 k−1 =Cn−kk
xossasiga binoan
un+un−1=1+∑k=1
s−1
Cn−k k +Cn−s−1 s−1 =1+∑k=1
s−1
Cn−k k +C2s−s−1 s−1 =
=1+∑k=1
s−1
Cn−k k +Cs−1s−1=Cn−k 0 +∑k=1
s−1
Cn−k k +Css=
=∑k=0
s−1
Cn−kk +Cn−ss =∑k=0
s
Cn−kk =un+1
tengliklarga ega bo‘lamiz .
n
toq son bo‘lganda ham, yuqoridagidek mulohazalar yuritib, un+1=un+un−1 (
n≥2
) tenglikning to‘g‘riligini ko‘rsatish mumkin. Demak, Fibonachchi qatorining
ta’rifiga asosan,
u1,u2,...,un,... ketma-ketligi Fibonachchi qatoridir.
Yuqorida ta’kidlanganidek,
un= ∑k=0
[n−12]
Cn−k−1 k tenglik Fibonachchi sonlari bilan
Paskal uchburchagi orasida bog‘lanishni ifodalayi. 1- shaklda tasvirlangan Paskal
uchburchagidagi shtrixli chiziqlar bo‘ylab joylashgan sonlar yig‘indisi Fibonachchi
sonlarini tashkil etadi.
12- x o s s a . Fibonachchi soni
un ( n∈N ) uchun
un= 1
√5[(
1+√5
2 )
n
−(
1− √5
2 )
n
]
tenglik o‘rinlidir .](https://docx.uz/documents/7a24074e-92fe-4a17-bfab-173d43d69571/page_15.png?v=1)
Paskal uchburchagi. nyuton binomi. binomial koeffitsiyent