\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+....+\frac{1}{9999}\)

=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)

=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

=\(1-\frac{1}{101}=\frac{100}{101}\)

tích trước đi đã!!!!!!!!!

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)

\(2\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\right)\)

\(2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(2\left(1-\frac{1}{101}\right)\)

\(2\times\frac{100}{101}\)

\(\frac{200}{101}\)

ta có:

\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)....\left(1+\frac{1}{9999}\right)\)

\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{10000}{9999}=\frac{2^2}{1.3}.\frac{3^2}{2.4}....\frac{100^2}{99.101}\)

\(A=\frac{\left(2.3.4.5....100\right)}{1.2.3.4....99}.\frac{\left(2.3.4...100\right)}{3.4.5..101}\)

\(A=\frac{100}{1}.\frac{2}{101}=\frac{200}{101}< \frac{202}{101}=2\)

\(\Rightarrow A< 2\)

nếu đúng k giúp mình nhé

Hi cám ơn bn nha

(1 + 1/3) × (1 + 1/8) × (1 + 1/15) × ... × (1 + 1/9999)

= 4/3 × 9/8 × 16/15 × ... × 10000/9999

= 2.2/1.3 × 3.3/2.4 × 4.4/3.5 × ... × 100.100/99.101

= 2.3.4...100/1.2.3...99 × 2.3.4...100/3.4.5...101

= 100 × 2/101

= 200/101

Ủng hộ mk nha ♡_♡

200/101

200/101