Đặt A = 1/2.6 + 1/6.10 + 1/10.14 + ..... + 1/102.106

=> 4A = 4/2.6 + 4/6.10 + 4/10.14 + ..... + 4/102.106

=> 4A = 1/2 - 1/6 + 1/6 - 1/10 + 1/10 - 1/14 + ... + 1/102 - 1/106

=> 4A = 1/2 - 1/106

=> 4A = 26/53

=> A = 13/106

~Study well~

#QASJ

\(\frac{1}{2.6}+\frac{1}{6.10}+...+\frac{1}{102.106}\)

\(=\frac{1}{4}.\left(\frac{4}{2.6}+\frac{4}{6.10}+...+\frac{4}{102.106}\right)\)

\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{102}-\frac{1}{106}\right)\)

\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{106}\right)\)

\(=\frac{1}{4}.\frac{26}{53}\)

\(=\frac{13}{106}\)

\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+...+\frac{1}{402.406}\)

4\(A=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{402}-\frac{1}{406}\)

4\(A=\frac{1}{6}-\frac{1}{406}\)

4\(A=\frac{100}{609}\)

\(\Rightarrow A=\frac{100}{609}:4\)\(=\frac{25}{609}\)

=1/6-1/10+1/10-1/14+1/14-1/18+...........+1/402-1/406

=1/6-1/406

12.T=2.6.12+6.10.12+10.14.12+...+102.106.12=

=2.6.(10+2)+6.10.(14-2)+10.14.(18-6)+...+102.106.(110-98)=

=2.2.6+2.6.10-2.6.10+6.10.14-6.10.14+10.14.18-...-98.102.106+102.106.110=

=2.2.6+102.106.110

\(\Rightarrow T=\dfrac{2.2.6+102.106.110}{12}=99112\)

bạn j ơi cho mình hỏi là 102 lấy ở đâu ja <3

Sửa đề: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)

Ta có: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)

\(=\dfrac{3}{4}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+\dfrac{4}{10\cdot14}+...+\dfrac{4}{26\cdot30}\right)\)

\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{26}-\dfrac{1}{30}\right)\)

\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{28}{60}\)

\(=\dfrac{21}{60}=\dfrac{7}{20}\)

\(B=\dfrac{1}{6.10}+\dfrac{1}{10.14}+...+\dfrac{1}{402.406}\\ 4B=\dfrac{4}{6.10}+\dfrac{4}{10.14}+...+\dfrac{4}{402.406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{402}-\dfrac{1}{406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}\\B=\dfrac{\dfrac{100}{609}}{4}=\dfrac{25}{609} \)

\(B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{ 1}{402}-\dfrac{1}{406}\)

\(=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}.\)

Đề sai bạn nhé.

Ta có: \(\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)

\(=\dfrac{3}{4}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+\dfrac{4}{10\cdot14}+...+\dfrac{4}{26\cdot30}\right)\)

\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{26}-\dfrac{1}{30}\right)\)

\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{7}{15}=\dfrac{21}{60}=\dfrac{7}{20}\)

a , 2/8 - 2/13 + 2/13 - 2/18 + 2/18 - 2/23 + ... + 2/253 - 2/258

= 2/8 - 2/258

= 125/516

b , 1/6 - 1/10 + 1/10 - 1/14 + 1/14 - 1/18 + ... + 1/402 - 1/406

= 1/6 - 1/406

= 100/609