Задача 61390 Найти сумму ряда....
Условие
Решение
[m]n^2-14n+48=(n-6)(n-8)[/m]
[m]\frac{2}{(n-6)(n-8)}=\frac{A}{n-6}+\frac{B}{n-8}[/m]
[m]2=A(n-8)+B(n-6)[/m]
n=6
[m]2=A(6-8)+B(6-6)[/m] ⇒ [m]A=-1[/m]
n=8
[m]2=A(8-8)+B(8-6)[/m] ⇒ [m]B=1[/m]
[m]\frac{2}{(n-6)(n-8)}=-\frac{1}{n-6}+\frac{1}{n-8}[/m]
[m]S_{n}=a_{9}+a_{10}+a_{11}...+a_{n-2}+a_{n-1}+a_{n}=[/m]
[m]=-\frac{1}{9-6}+\frac{1}{9-8}-\frac{1}{10-6}+\frac{1}{10-8}-\frac{1}{11-6}+\frac{1}{11-8}-\frac{1}{12-6}+\frac{1}{12-8}-\frac{1}{13-6}+\frac{1}{13-8}+...-\frac{1}{n-2-6}+\frac{1}{n-2-8}-\frac{1}{n-1-6}+\frac{1}{n-1-8}-\frac{1}{n-6}+\frac{1}{n-8}[/m]
[m]=-\frac{1}{3}+1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\frac{1}{5}+...-\frac{1}{n-8}+\frac{1}{n-10}-\frac{1}{n-7}+\frac{1}{n-9}-\frac{1}{n-6}+\frac{1}{n-8}[/m]
[m]-\frac{1}{3}+\frac{1}{3}=0[/m]
[m]-\frac{1}{4}+\frac{1}{4}=0[/m]
[m]-\frac{1}{5}+\frac{1}{5}=0[/m]
[m]-\frac{1}{n-8}+\frac{1}{n-8}=0[/m]
⇒
Останутся только
[m]=+1+\frac{1}{2}-\frac{1}{n-7}-\frac{1}{n-6}[/m]
[m]S=lim_{n → ∞ }S_{n}=1+\frac{1}{2}-0-0=\frac{3}{2}[/m]