Задача 64637 Найти сумму ряда...
Условие
математика ВУЗ
109
Решение
★
[m]n^2-12n+35=(n-5)(n-7)[/m]
[m]\frac{36}{n^2-12n+35}=36\cdot \frac{1}{2}(\frac{1}{n-7}-\frac{1}{n-5})[/m]
[m]\frac{36}{n^2-12n+35}=18\cdot (\frac{1}{n-7}-\frac{1}{n-5})[/m]
Тогда
[m] S_{n}=18\cdot (\frac{1}{9-7}-\frac{1}{9-5})+18\cdot (\frac{1}{10-7}-\frac{1}{10-5})+18\cdot (\frac{1}{11-7}-\frac{1}{11-5})+[/m]
[m]+18\cdot (\frac{1}{n-1-7}-\frac{1}{n-1-5})+18\cdot (\frac{1}{n-7}-\frac{1}{n-5})=18\cdot (\frac{1}{2}+\frac{1}{3}-\frac{1}{n-6}-\frac{1}{n-5})[/m]
[m]S=lim_{n → ∞ } S_{n} =lim_{n → ∞ }18\cdot (\frac{1}{2}+\frac{1}{3}-\frac{1}{n-6}-\frac{1}{n-5})=18\cdot\frac{5}{6}=15[/m]