Đáp án:
Giải thích các bước giải:
Ta có : $\frac{n² - 1}{n²} = \frac{n - 1}{n}.\frac{n + 1}{n}$ nên :
$A = \frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{9603}{9604}.\frac{9800}{9801}$
$= \frac{1}{2}.\frac{3}{2}.\frac{2}{3}.\frac{4}{3}.\frac{3}{4}.\frac{5}{4}.\frac{4}{5}.\frac{6}{5}...\frac{97}{98}.\frac{99}{98}.\frac{98}{99}.\frac{100}{99} = \frac{1}{2}.\frac{100}{99} = \frac{50}{99}$
$C = 1.2 + 2.3 + 3.4 + ..+ 97.98 + 98.99$
$⇒ 3C = 1.2.3 + 2.3.3 + 3.4.3 +...+ 97.98.3 + 98.99.3 = 1.2.3 + 2.3.(4 - 1) + 3.4.(5- 2) +...+ 97.98.(99 - 96) + 98.99.(100 - 97) = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ 97.98.99 - 96.97.98 + 98.99.100 - 97.98.99 = 98.99.100$
$ B = \frac{1.2 + 2.3 + 3.4 + ..+ 97.98 + 98.99}{98.99.100} = \frac{C}{98.99.100} = \frac{1}{3}$
$⇒ A > B$
