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Tushunganingizni tekshiring
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matematika N.Yuldasheva mustaqil ish
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- 2) Ixtiyoriy � n n had raqami uchun �(�−1) a ( n −1)a, left parenthesis, n, minus, 1, right parenthesis nimani ifodalaydi
- Arifmetik progressiyaning rekursiv formulasi
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: Progressiyaning birinchi hadi 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:
Qoyil! Bu formula bizga 3, 5, 7, ... progressiya kabi progressiyani berdi. Download 337.13 Kb. Do'stlaringiz bilan baham: |
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