Question

Найдите номер наибольшего члена последовательности , если an=10n/(100+n^2)

Answer 1

$\frac{10n}{100+n^2} \geq \frac{10(n+1)}{100+(n+1)^2}\\ 10n(100+(n+1)^2) \geq 10(n+1)(100+n^2)\\ n(100+(n+1)^2) \geq (n+1)(100+n^2)\\ 100n+n(n+1)^2 \geq 100n+100+n^3+n^2\\ n(n+1)^2 \geq 100+n^3+n^2\\ n(n^2+2n+1) \geq 100+n^3+n^2\\ n^3+2n^2+n \geq 100+n^3+n^2\\ n^2+n \geq 100\\ n^2+n+0,25\geq 100,25\\ (n+0,5)^2\geq 100,25\\ n+0,510\\n9,5\\n=10$
$a_1= \frac{10}{101}\\ a_2= \frac{20}{104}\\ a_3= \frac{30}{109} \\...\\a_8= \frac{80}{164}\\ a_9= \frac{90}{181}\\ a_{10}= \frac{100}{200}=0,5\\ a_{11}= \frac{110}{221}$
ответ: 10.