Đáp án:
$\frac{98}{15}$
Giải thích các bước giải:
A = $\frac{2}{3}$ + $\frac{14}{15}$ + $\frac{34}{35}$ + $\frac{62}{63}$ + $\frac{98}{99}$ + $\frac{142}{143}$ + $\frac{194}{195}$
= (1 - $\frac{1}{3}$) + (1 - $\frac{1}{15}$) + (1 - $\frac{1}{35}$) + (1 - $\frac{1}{63}$) + (1 - $\frac{1}{99}$) + (1 - $\frac{1}{143}$) + (1 - $\frac{1}{195}$)
= (1 + 1 + 1 + 1 + 1 + 1 + 1) - ($\frac{1}{3}$ + $\frac{1}{15}$ + $\frac{1}{35}$ + $\frac{1}{63}$ + $\frac{1}{99}$ + $\frac{1}{143}$ + $\frac{1}{195}$)
= 7 - ($\frac{1}{1.3}$ + $\frac{1}{3.5}$ + $\frac{1}{5.7}$ + $\frac{1}{7.9}$ + $\frac{1}{9.11}$ + $\frac{1}{11.13}$ + $\frac{1}{13.15}$)
Đặt S = $\frac{1}{1.3}$ + $\frac{1}{3.5}$ + $\frac{1}{5.7}$ + $\frac{1}{7.9}$ + $\frac{1}{9.11}$ + $\frac{1}{11.13}$ + $\frac{1}{13.15}$
⇒ A = 7 - S
Ta có: 2S = 2.($\frac{1}{1.3}$ + $\frac{1}{3.5}$ + $\frac{1}{5.7}$ + $\frac{1}{7.9}$ + $\frac{1}{9.11}$ + $\frac{1}{11.13}$ + $\frac{1}{13.15}$)
= $\frac{2}{1.3}$ + $\frac{2}{3.5}$ + $\frac{2}{5.7}$ + $\frac{2}{7.9}$ + $\frac{2}{9.11}$ + $\frac{2}{11.13}$ + $\frac{2}{13.15}$
= 1 - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{5}$ + $\frac{1}{5}$ - $\frac{1}{7}$ + $\frac{1}{7}$ - $\frac{1}{9}$ + $\frac{1}{9}$ - $\frac{1}{11}$ + $\frac{1}{11}$ - $\frac{1}{13}$ + $\frac{1}{13}$ - $\frac{1}{15}$
= 1 - $\frac{1}{15}$ = $\frac{14}{15}$
⇒ S = $\frac{14}{15}$ : 2 = $\frac{7}{15}$
⇒ A = 7 - S = 7 - $\frac{7}{15}$ = $\frac{98}{15}$
Vậy A = $\frac{2}{3}$ + $\frac{14}{15}$ + $\frac{34}{35}$ + $\frac{62}{63}$ + $\frac{98}{99}$ + $\frac{142}{143}$ + $\frac{194}{195}$ = $\frac{98}{15}$
Chúc bn học tốt!