Sayxun Team vs Guliston A cademy \( \begin{array}{l}\text { Azizbek Nuraliyev } \\ \text { Sanjar Rahimj onov } \\ \text { [2,2 ball] }\end{array} \) 21. Hisoblang: \( \frac{1}{\log _{2} 15!}+\frac{1}{\log _{3} 15!}+\frac{1}{\log _{4} 15!}+\cdots+\frac{1}{\log _{15} 15!} \) A) 1

Berilgan ifoda: \[ S = \frac{1}{\log_{2} 15!} + \frac{1}{\log_{3} 15!} + \frac{1}{\log_{4} 15!} + \cdots + \frac{1}{\log_{15} 15!} \] Bu ifodani soddalashtirish uchun, logarifmning o'zgarish xususiyatidan foydalanamiz: \[ \log_{b} a = \frac{\log_{k} a}{\log_{k} b} \] Bu yerda \( k \) istalgan musbat son bo'lishi mumkin. Biz \( k = 15! \) ni tanlaymiz. Shunday qilib, har bir logarifmni quyidagicha yozishimiz mumkin: \[ \log_{b} 15! = \frac{\log 15!}{\log b} \] Shu sababli, ifodamiz quyidagicha bo'ladi: \[ S = \sum_{b=2}^{15} \frac{1}{\frac{\log 15!}{\log b}} = \sum_{b=2}^{15} \frac{\log b}{\log 15!} \] Bu ifodani soddalashtirsak: \[ S = \frac{1}{\log 15!} \sum_{b=2}^{15} \log b \] Logarifmning yig'indisi quyidagicha ifodalanadi: \[ \sum_{b=2}^{15} \log b = \log(2 \cdot 3 \cdot 4 \cdots \cdot 15) = \log(15!) \] Shunday qilib, ifodamiz quyidagicha bo'ladi: \[ S = \frac{1}{\log 15!} \cdot \log(15!) = 1 \] Natijada, berilgan ifoda: \[ \frac{1}{\log_{2} 15!} + \frac{1}{\log_{3} 15!} + \frac{1}{\log_{4} 15!} + \cdots + \frac{1}{\log_{15} 15!} = 1 \] Shunday qilib, to'g'ri javob: **A) 1**