\(a,=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

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

\(=\frac{1}{2}-0-0-0-...-0-\frac{1}{8}\)

\(=\frac{1}{2}-\frac{1}{8}\)

\(=\frac{4}{8}-\frac{1}{8}\)

\(=\frac{3}{8}\)

\(b,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{49}+\frac{1}{49}-\frac{1}{16}\)

\(=1-0-0-0-...-0-\frac{1}{16}\)

\(=1-\frac{1}{16}\)

\(=\frac{15}{16}\)

\(c,\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\left(1-0-0-0-...-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\frac{50}{51}\)

\(=\frac{25}{17}\)

\(d,\)giống câu a tự làm nha mỏi tay quá.

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}.\)

=> \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)

=> \(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

\(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{49.52}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{49}-\frac{1}{52}\)

=> \(B=\frac{1}{4}-\frac{1}{52}=\frac{24}{104}=\frac{1}{26}\)

1/2*3+1/3*4+1/4*5+...+1/7*8

1/2-1/3+1/3-1/4+1/4-1/5-...-1/8

1/2-1/8=3/8

1/4-1/7+1/7-1/10+1/10-1/13-...-1/52                                   49/52 bạn nhé

1/4-1/52=3/13

câu này mình gọi nó là S

ta có S:2=2/1*3+2/3*5+...+2/49*51

1/1-1/3+1/3-1/5+...+1/49-1/51

1/1-1/51=50/51

S=50/51*2=100/51

1/100-1/101+1/101-1/102+1/102-1/103+...+1/2010-1/2011

1/100-1/2011

bạn tích đi nhé mình còn phải đi học bạn k cho mình nhé

A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102

A=1-1/102=102/102-1/102=101/102

ý b thì chờ mình tí tìm cách lập luận đã nhé

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)

\(A=1-\frac{1}{102}\)

\(A=\frac{101}{102}\)

B=1/1.2-1/2.3-1/3.4-1/4.5-.......1/100.101-1/101.102

B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......+1/100-1/101+1/101-1/102

B=1-1/102

a)    \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

      \(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)(áp dụng quy tắc dấu ngoặt )

\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^8}\)

\(3A-A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^7}-\frac{1}{3^7}\right)-\frac{1}{3^8}\)

\(\Rightarrow2A=1+0+0...+0-\frac{1}{3^8}\)

     \(2A=1-\frac{1}{3^8}\)

     \(2A=\frac{3^8-1}{3^8}\)

     \(A=\frac{3^8-1}{3^8}\div2=\frac{3^8-1}{3^8}.\frac{1}{2}=\frac{3^8-1}{3^8.2}\)

b)   \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}\)

     \(\Rightarrow B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)(áp dụng quy tắc dấu ngoặt )

      \(B=\frac{1}{1}-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{100}-\frac{1}{100}\right)-\frac{1}{101}\)

     \(B=\frac{1}{1}-0-0-0...-0-\frac{1}{101}\)

      \(B=\frac{1}{1}-\frac{1}{101}\)

      \(B=\frac{100}{101}\)

B=1/1-1/2+1/2-1/3+1/3-1/4+...+

mik kg hiểu bạn đang làm cái gì cả.

\(D=\frac{\left(-1\right).\left(-1\right)}{1.2}.\frac{\left(-2\right).\left(-2\right)}{2.3}...\frac{\left(-101\right).\left(-101\right)}{101.102}\)

\(=\frac{\left(-1\right)\left(-1\right)\left(-2\right)\left(-2\right)...\left(-101\right)\left(-101\right)}{1.2.2.3...101.102}\)

\(=\frac{\left[\left(-1\right)\left(-2\right)...\left(-101\right)\right].\left[\left(-1\right).\left(-2\right)...\left(-101\right)\right]}{\left(1.2...101\right).\left(2.3...102\right)}\)

\(=\left(-1\right).\frac{-1}{102}\)

\(=\frac{1}{102}\)

Vì \(\frac{1}{102}>\frac{-1}{100}\)

Vậy\(D>\frac{-1}{100}\)

Đặt \(A=\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(\Rightarrow A< \left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...+\frac{1}{100.101}\right)+\left(\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}\right)\)

\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{101}< \frac{1}{2}\)

\(\Rightarrow A< \frac{1}{2}\)

Vậy \(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...+\frac{1}{100.101}< 2\) (đpcm)

Mai ơi, bài này thầy dạy hôm chiều cậu nghỉ đó