S=1/4+1/9+1/16+1/25+1/36+1/49+1/64+1/81=1-1/81=1/81

80/81 là đúng

\(\left(1-\frac{4}{36}\right)\left(1-\frac{4}{49}\right)\cdot\ldots\cdot\left(1-\frac{4}{625}\right)\)

\(=\left(1-\frac26\right)\left(1-\frac27\right)\cdot\ldots\cdot\left(1-\frac{2}{25}\right)\left(1+\frac26\right)\left(1+\frac27\right)\cdot\ldots\cdot\left(1+\frac{2}{25}\right)\)

\(=\frac46\cdot\frac57\cdot\ldots\cdot\frac{23}{25}\cdot\frac86\cdot\frac97\cdot\ldots\cdot\frac{27}{25}\)

\(=\left(\frac46\cdot\frac68\cdot\ldots\cdot\frac{22}{24}\right)\left(\frac57\cdot\frac79\cdot\ldots\cdot\frac{23}{25}\right)\cdot\left(\frac86\cdot\frac{10}{8}\cdot\ldots\cdot\frac{26}{24}\right)\cdot\left(\frac97\cdot\frac{11}{9}\cdot\ldots\cdot\frac{27}{25}\right)\)

\(=\frac{4}{24}\cdot\frac{5}{25}\cdot\frac{26}{6}\cdot\frac{27}{7}=\frac14\cdot\frac15\cdot\frac{13}{3}\cdot\frac{27}{7}=\frac{13\cdot9}{7\cdot4\cdot5}\)

\(\frac{1}{23}-\frac{4}{23\cdot25}\)

\(=\frac{25-4}{25\cdot23}=\frac{21}{25\cdot23}\)

Ta có: \(C=\left(\frac{1}{23}-\frac{4}{23\cdot25}\right)\cdot\left(\left(1-\frac{4}{36}\right)\left(1-\frac{4}{49}\right)\cdot\ldots\cdot\left(1-\frac{4}{625}\right)\right)\)

\(=\frac{7\cdot3}{5\cdot5\cdot23}\cdot\frac{13\cdot9}{7\cdot4\cdot5}=\frac{3\cdot13\cdot9}{5\cdot5\cdot23\cdot4\cdot5}=\frac{27\cdot13}{125\cdot4\cdot23}=\frac{351}{11500}\)

a) Ta có 

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{8.8}\)

Mà \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{8.8}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)

                                                      \(=1-\frac{1}{8}\)

                                                       \(=\frac{7}{8}<1\)

Vì \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{8.8}<\frac{7}{8}<1\)

nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}<1\)

D

thank

A=1/2²+1/3²+...+1/9²

Ta có:1/2²>1/2.3,1/3²>1/3.4,...,1/9²>1/9.10

⇒A>1/2.3+1/3.4+...+1/9.10

A>1/2-1/3+1/3-1/4+...+1/9-1/10

A>1/2-1/10

A>2/5(đpcm)

Ta có: A = 1/4 + 1/9 + 1/16 + 1/25 +1/36 + 1/49 + 1/64 + 1/81

Vì 1/2²>1/2.3,1/3²>1/3.4,...,1/9²>1/9.10

=>A>1/2.3+1/3.4+...+1/9.10

=>A>1/2-1/3+1/3-1/4+...+1/9-1/10

=>A>1/2-1/10

=>A>2/5

Chọn A

A