University of business and science

Tushunganingizni tekshiring
Endi rekursiv formuladan foydalanib, progressiya hadlarini toping.
�(�)a(n)a, left parenthesis, n, right parenthesisdan 3, 5, 7, ... progressiyaning �-n-n, start text, negative, end texthadini ifodalashda foydalangan boʻlsak, boshqa harflardan qolgan progressiyalarni ifodalashda foydalanishimiz mumkin. Masalan, �(�)b(n)b, left parenthesis, n, right parenthesis, �(�)c(n)c, left parenthesis, n, right parenthesis yoki �(�)d(n)d, left parenthesis, n, right parenthesis.
3) Progressiya haqida berilganlar yordamida �(4)b(4)b, left parenthesis, 4, right parenthesis ni toping. {�(1)=−5�(�)=�(�−1)+9⎩⎪⎪⎨⎪⎪⎧​b(1)=−5b(n)=b(n−1)+9​
�(4)=b(4)=b, left parenthesis, 4, right parenthesis, equals 

Tekshirish
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4) Progressiya haqida berilganlar yordamida �(3)c(3)c, left parenthesis, 3, right parenthesis ni toping. {�(1)=20�(�)=�(�−1)−17⎩⎪⎪⎨⎪⎪⎧​c(1)=20c(n)=c(n−1)−17​
�(3)=c(3)=c, left parenthesis, 3, right parenthesis, equals 

Tekshirish
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5) Progressiya haqida berilganlar yordamida �(4)d(4)d, left parenthesis, 4, right parenthesis ni toping. {�(1)=2�(�)=�(�−1)+0.4⎩⎪⎪⎨⎪⎪⎧​d(1)=2d(n)=d(n−1)+0.4​
�(5)=d(5)=d, left parenthesis, 5, right parenthesis, equals 

Tekshirish
[Yordam kerak!]
Arifmetik progressiyaning oshkor formulasi
Quyida 3, 5, 7, ... uchun oshkor formula berilgan:
�(�)=3+2(�−1)a(n)=3+2(n−1)a, left parenthesis, n, right parenthesis, equals, 3, plus, 2, left parenthesis, n, minus, 1, right parenthesis
Bu formula yordamida biz shunchaki had raqamini formulaga qoʻyib, kerakli hadni topishimiz mumkin.
Masalan, beshinchi hadni topish uchun oshkor formulaga �=5n=5n, equals, 5 ni qoʻyishimiz kerak.
�(5)=3+2(5−1)=3+2⋅4=3+8=11a(5)​=3+2(5−1)=3+2⋅4=3+8=11​
Ajoyib! Biz xuddi oʻsha javobni oldik!