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Tushunganingizni tekshiring


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Bog'liq
matematika N.Yuldasheva mustaqil ish

Tushunganingizni tekshiring
1) 3, 5, 7, ... progreessiyada �(4)a(4)a, left parenthesis, 4, right parenthesis ni toping.
�(4)=a(4)=a, left parenthesis, 4, right parenthesis, equals 

Tekshirish
[Yordam kerak!]
2) Ixtiyoriy �nn had raqami uchun �(�−1)a(n−1)a, left parenthesis, n, minus, 1, right parenthesis nimani ifodalaydi?
Bitta javobni tanlang:
Bitta javobni tanlang:

  • (Tanlov A)

111 dan �-n-n, start text, negative, end texthadni ayirish
A
111 dan �-n-n, start text, negative, end texthadni ayirish

  • (Tanlov B)

�-n-n, start text, negative, end texthaddan oldingi had
B
�-n-n, start text, negative, end texthaddan oldingi had
Tekshirish
[Yordam kerak!]
Arifmetik progressiyaning rekursiv formulasi
Rekursiv formula bizga ikkita qism haqida maʼlumot beradi:

  1. Progressiyaning birinchi hadi

  2. Oldingi haddan foydalanib progressiyaning ixtiyoriy hadini topish uchun namuna

Quyida 3, 5, 7, ... progressiya uchun rekursiv formula keltirilgan.
{�(1)=3←Birinchi had - uch.�(�)=�(�−1)+2←Oldingi hadga ikkini qoʻshing.⎩⎪⎪⎨⎪⎪⎧​a(1)=3a(n)=a(n−1)+2​←Birinchi had - uch.←Oldingi hadga ikkini qoʻshing.​
Masalan, beshinchi hadni topish uchun biz progressiyani had bo‘yicha kengaytiramiz:

(�)a(n)a, left parenthesis, n, right parenthesis

=�(� ⁣− ⁣ ⁣1)+2=a(n−1)+2equals, a, left parenthesis, n, minus, 1, right parenthesis, plus, 2










�(1)a(1)a, left parenthesis, 1, right parenthesis







=3=3equals, start color #11accd, 3, end color #11accd




�(2)a(2)a, left parenthesis, 2, right parenthesis

=�(1)+2=a(1)+2equals, a, left parenthesis, 1, right parenthesis, plus, 2

=3+2=3+2equals, start color #11accd, 3, end color #11accd, plus, 2

=5=5equals, start color #aa87ff, 5, end color #aa87ff




�(3)a(3)a, left parenthesis, 3, right parenthesis

=�(2)+2=a(2)+2equals, a, left parenthesis, 2, right parenthesis, plus, 2

=5+2=5+2equals, start color #aa87ff, 5, end color #aa87ff, plus, 2

=7=7equals, start color #1fab54, 7, end color #1fab54




�(4)a(4)a, left parenthesis, 4, right parenthesis

=�(3)+2=a(3)+2equals, a, left parenthesis, 3, right parenthesis, plus, 2

=7+2=7+2equals, start color #1fab54, 7, end color #1fab54, plus, 2

=9=9equals, start color #e07d10, 9, end color #e07d10




�(5)a(5)a, left parenthesis, 5, right parenthesis

=�(4)+2=a(4)+2equals, a, left parenthesis, 4, right parenthesis, plus, 2

=9+2=9+2equals, start color #e07d10, 9, end color #e07d10, plus, 2

=11=11equals, 11




Qoyil! Bu formula bizga 3, 5, 7, ... progressiya kabi progressiyani berdi.

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