Geometrik progressiyaning dastlabki n ta hadining yig’indisi

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Dars mavzusi :
« Geometrik progressiyaning
dastlabki n ta hadining yig’indisi »

1-misol: Ushbu yig’indini toping.

Tenglikning ikkala qismini 3 ga ko’paytiramiz. 5 4 3 2
3 3 3 3 3 1       S
6 5 4 3 2
3 3 3 3 3 3 3       S
Tengliklarni bunday yozib chiqamiz.
) 3 3 3 3 3( 1
5 4 3 2
      S
6 5 4 3 2
3)33333(3 S
Qavsdagi ifodalar bir xil. Shuning uchun pastdagi
tenglikdan yuqoridagi tenglikni ayirib , hosil qilamiz.
1 3 3
6
   S S
1 3 2
6
  S
364
2
1 729
2
1 3
6




 S
5 4 3 2
3 3 3 3 3 1       S
6 5 4 3 2
3 3 3 3 3 3 3       S
) 3 3 3 3 3( 1
5 4 3 2
      S
6 5 4 3 2
3 ) 3 3 3 3 3( 3       S
1 3 3
6
   S S
1 3 2
6
  S
364
2
1 729
2
1 3
6




 S

Maxraji bo’lgan ixtiyoriy geometrik
progressiyaning qaraymiz. Shu progressiyaning n ta
hadining yig’indisini topaylik,
Teorema. Maxraji bo’lgan geometrik
progressiyaning dastlabki n ta hadining yig’indisi
quyidagiga teng.
1
1
2
1 1 1 ...

    
n
n q b q b q b b S
q
q b
S
n
n



1
) 1(1
1  q n b b b b ..., , , , 3 2 1
1  q
indisi yig hadining n dastlabki S n ' 
hadi birinchi b  1
xraji ma q  1
12
111
...      n
n q b q b q b b S
q
q b
S
n
n



1
) 1(1
1  q n b b b b ..., , , , 3 2 1
1  q
indisi yig hadining n dastlabki S n ' 
hadi birinchi b  1
xraji ma q 

Isbot: Tenglikning ikkala
qismini q ga ko’paytiramiz.
Yuqoridagi ikkala tenglikdan bir xil qo’shiluvchilarni
ajratib yozib chiqamiz.1
1
2
1 1 1 ...

    
n
n q b q b q b b S
n
n q b q b q b q b qS 1
3
1
2
1 1 ...     
) ... (
1
1
2
1 1 1

    
n
n q b q b q b b S
n n
n q b q b q b q b q b qS 1
1
1
3
1
2
1 1 ) ... (      

Qavs ichida turgan ifodalar teng. Shuning uchun
yuqoridagi tenglikdan pastdagisini ayirib, hosil qilamiz.
n
n n q b b qS S 1 1   
) 1( ) 1( 1
n
n q b q S   
q
q b
S
n
n



1
) 1(1 1
12
111
...      n
n q b q b q b b S
n
n q b q b q b q b qS 1
3
1
2
1 1 ...     
) ... (
1
1
2
1 1 1

    
n
n q b q b q b b S
n n
n q b q b q b q b q b qS 1
1
1
3
1
2
1 1 ) ... (      

n
n n q b b qS S 1 1   
) 1( ) 1(
1 n
n q b q S   
q
q b
S
n
n



1
) 1(1

2-misol: geometrik progressiya dastlabki beshta
hadining yig’indisini hisoblang. ...,
3
2
, 2 , 6
Bu progressiyada
31
,6
1  qb
27
242
243 2
3 242 6
3
2
243
1
1 6
3
1
1
3
1
1 6
1
) 1(
5
5
1
5 

 







 














 




q
q b
S
...,
3
2
, 2 , 6
3
1
, 6 1   q b
27
242
243 2
3 242 6
3
2
243
1
1 6
3
1
1
3
1
1 6
1
) 1(
5
5
1
5 

 







 














 




q
q b
S

Mustahkamlash uchun masala
3-masala. Shaxmatning kelib
chiqishi haqida afsona:

Shaxmat o’yini Hindistonda o’ylab chiqarilgan
bo’lib, Sheram ismli hind shohi shu o’yin bilan
tanishib chiqqandan keyin uni o’ylab topgan o’zining
fuqoralaridan biri ekanligini bilib unga o’z qo’li bilan
mukofat bermoqchi bo’ladi. Shaxmatni o’ylab
chiqargan kishi Seta nomli donishmand shoh
huzuriga kelib tazim bajo keltiradi.
Shoh: - O’ylab chiqargan ajoyib o’yining
uchun senga katta mukofat bilan xursand
qilmoqchiman,-dedi.

Tilagingni bajarishda sendan hech narsani
ayamayman. –Lutfi keng shohim o’ylab ko’ray keyin
javobini aytaman debdi. Ertasi kuni Seta shoh oldiga
kelib,-shaxmat taxtasining
birinchi katagi uchun 1 dona,
ikkinchisi uchun 2 dona,
uchinchisi uchun 4 dona,
to’rtinchisi uchun 8 dona,
beshinchisi uchun 16 dona,
oltinchisi uchun 32 dona,
Bas etar debdi shoh zarda bilan.
Navkarlarim sen so’ragan narsani olib
chiqib beradi debdi. Ertasiga hisobchilar
hisoblab chiqib. Bu son shu qadar kattaki,
bu tilakni bajarishga qodir emasmiz debdi.

Bu son 18 446 744 073 709 551 615
18 kvintillion 446 kvadrilion 774 trillion 73 milliard
709 million 551 ming 615 dona
Birinchi hadi 1, ikkinchi hadi 2 bo’lgan geometrik
progressiya 
1 2
2 1
2 1 1
2 ... 2 2 1
64
64
63 3 2
64  


      S
 
1 2
2 1
2 1 1
2 ... 2 2 1
64
64
63 3 2
64  


      S

Mustahkamlash uchun:
“ Bilimdonlar maydoni” o’yini

Mustahkamlash uchun savollar:
1. Geometrik progreesiya dastlabki n ta hadining yig’indisi
qanday topiladi?
2. Geometrik progressiya n-hadini topish formulasini
ayting.
3. b
n va q ning farqi nimada?
4. Guruh nomlariga haqida qisqacha ta’rif bering

6. Ushbu afsonadagi maxrajini ayting 5. Shaxmat afsonasidagi keltirilgan misolda qaysi
progressiyaga oidligini tushundingiz

Dars yakuni
Geometrik progressiyaning dastlabki n ta hadining
yig’indisining formulasi va unga oid misollar yechish1. Bugungi darsda nimalarni o’rgandik

Uyga vazifa
33-§.
№ 440(2,4, 6) ;
№ 441(2) ;
№ 442 ( 2 ).

Foydalanilgan adabiyotlar:
1. Algebra: Umumiy o’rta ta’lim maktablarining 9 -sinfi uchun darslik/
Sh.A . Alimov , O . R . Xolmuhammedov , M . A . Mirzaahmedov 3-nashri.-T.:
“ O’qituvchi” NMIU, 20 14 .- 240 b
2. 9 -sinfda Algebra O’qituvchilar uchun uslubiy qo’llanma/
Sh.A . Alimov , O . R . Xolmuhammedov , M . A . Mirzaahmedov 3-nashri.-T.:
“ O’qituvchi” NMIU, 200 9 .-207 b

Dars mavzusi : « Geometrik progressiyaning dastlabki n ta hadining yig’indisi »

1-misol: Ushbu yig’indini toping. Tenglikning ikkala qismini 3 ga ko’paytiramiz. 5 4 3 2 3 3 3 3 3 1       S 6 5 4 3 2 3 3 3 3 3 3 3       S Tengliklarni bunday yozib chiqamiz. ) 3 3 3 3 3( 1 5 4 3 2       S 6 5 4 3 2 3)33333(3 S Qavsdagi ifodalar bir xil. Shuning uchun pastdagi tenglikdan yuqoridagi tenglikni ayirib , hosil qilamiz. 1 3 3 6    S S 1 3 2 6   S 364 2 1 729 2 1 3 6      S 5 4 3 2 3 3 3 3 3 1       S 6 5 4 3 2 3 3 3 3 3 3 3       S ) 3 3 3 3 3( 1 5 4 3 2       S 6 5 4 3 2 3 ) 3 3 3 3 3( 3       S 1 3 3 6    S S 1 3 2 6   S 364 2 1 729 2 1 3 6      S

Maxraji bo’lgan ixtiyoriy geometrik progressiyaning qaraymiz. Shu progressiyaning n ta hadining yig’indisini topaylik, Teorema. Maxraji bo’lgan geometrik progressiyaning dastlabki n ta hadining yig’indisi quyidagiga teng. 1 1 2 1 1 1 ...       n n q b q b q b b S q q b S n n    1 ) 1(1 1  q n b b b b ..., , , , 3 2 1 1  q indisi yig hadining n dastlabki S n '  hadi birinchi b  1 xraji ma q  1 12 111 ...      n n q b q b q b b S q q b S n n    1 ) 1(1 1  q n b b b b ..., , , , 3 2 1 1  q indisi yig hadining n dastlabki S n '  hadi birinchi b  1 xraji ma q 

Isbot: Tenglikning ikkala qismini q ga ko’paytiramiz. Yuqoridagi ikkala tenglikdan bir xil qo’shiluvchilarni ajratib yozib chiqamiz.1 1 2 1 1 1 ...       n n q b q b q b b S n n q b q b q b q b qS 1 3 1 2 1 1 ...      ) ... ( 1 1 2 1 1 1       n n q b q b q b b S n n n q b q b q b q b q b qS 1 1 1 3 1 2 1 1 ) ... (        Qavs ichida turgan ifodalar teng. Shuning uchun yuqoridagi tenglikdan pastdagisini ayirib, hosil qilamiz. n n n q b b qS S 1 1    ) 1( ) 1( 1 n n q b q S    q q b S n n    1 ) 1(1 1 12 111 ...      n n q b q b q b b S n n q b q b q b q b qS 1 3 1 2 1 1 ...      ) ... ( 1 1 2 1 1 1       n n q b q b q b b S n n n q b q b q b q b q b qS 1 1 1 3 1 2 1 1 ) ... (        n n n q b b qS S 1 1    ) 1( ) 1( 1 n n q b q S    q q b S n n    1 ) 1(1

2-misol: geometrik progressiya dastlabki beshta hadining yig’indisini hisoblang. ..., 3 2 , 2 , 6 Bu progressiyada 31 ,6 1  qb 27 242 243 2 3 242 6 3 2 243 1 1 6 3 1 1 3 1 1 6 1 ) 1( 5 5 1 5                                  q q b S ..., 3 2 , 2 , 6 3 1 , 6 1   q b 27 242 243 2 3 242 6 3 2 243 1 1 6 3 1 1 3 1 1 6 1 ) 1( 5 5 1 5                                  q q b S

Geometrik progressiyaning dastlabki n ta hadining yig’indisi

20.11.2024

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1595.1357421875 KB

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