Question

a 1 ​ ,a 2 ​ ,a 3 ​ ,⋯ kamaymaydigan natural sonlar ketma-ketligi. Ixtiyoriy natural m≥1 uchun b_{m}b m ​ ketma-ketlikni b_{m} =\min \left\{n:a_{n} \ge m\right\}b m ​ =min{n:a n ​ ≥m} ko’rinishida aniqlaylik. Agar a_{221} =1545a 221 ​ =1545 bo’lsa, u holda quyidagi ifodaning eng katta qiymatini toping a_{1} +a_{2} +\cdots +a_{221} +b_{1} +b_{2} +\cdots +b_{1545}a 1 ​ +a 2 ​ +⋯+a 221 ​ +b 1 ​ +b 2 ​ +⋯+b 1545 ​
60 bal beraman ​