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

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

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

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

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

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

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

Đặt \(B=\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{102.106}\)

\(\Leftrightarrow4B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)

\(\Leftrightarrow4B=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)

\(\Leftrightarrow4B=\frac{1}{2}-\frac{1}{106}=\frac{26}{53}\)

\(\Leftrightarrow B=\frac{26}{53}:4=\frac{13}{106}\)

Vậy ...................

~ Hok tốt ~

\(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)

\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)

Vậy \(A=\frac{504}{1009}.\)

\(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)

\(=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)

\(=\frac{1}{2}-\frac{1}{106}=\frac{26}{53}\)

Vậy \(B=\frac{26}{53}.\)

Bài làm:

a) \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)

\(A=\frac{1}{2}-\frac{1}{2018}\)

\(A=\frac{504}{1009}\)

b) \(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)

\(B=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)

\(B=\frac{1}{2}-\frac{1}{106}\)

\(B=\frac{26}{53}\)

\(A=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{2015.2018}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{2015}-\frac{1}{2018}\)

\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)

\(B=\frac{4}{2.6}+\frac{4}{6.10}+...+\frac{4}{102.106}=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{102}-\frac{1}{106}=\frac{1}{2}-\frac{1}{106}\)

\(=\frac{26}{53}\)

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

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

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

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

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

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

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

=3/12+3/60+3/140

=>1/4+1/20+3/140

=>4+20+3/140

=>3363/140

\(S=\dfrac{2}{2\cdot6}+\dfrac{2}{6\cdot10}+...+\dfrac{2}{96\cdot100}\\ =\dfrac{1}{2}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+...+\dfrac{4}{96\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+...+\dfrac{1}{96}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=\dfrac{1}{2}\cdot\dfrac{49}{100}\\ =\dfrac{49}{100}\)

Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

Bài 6 :

\(C=1^2+2^2+...+100^2=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}=\dfrac{100.101.201}{6}=338350\)

Bài 9 :

\(S=1^2+2^2+3^2+...+99^2=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}=\dfrac{99.100.199}{6}=328350\)

   \(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)

=  \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))

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

= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))

= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)

= \(\dfrac{9}{28}\)

B = \(\dfrac{4}{1.3.5}\) + \(\dfrac{4}{3.5.7}\) + \(\dfrac{4}{5.7.9}\)

B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{3.5}\) + \(\dfrac{1}{3.5}\) - \(\dfrac{1}{5.7}\) + \(\dfrac{1}{5.7}\) - \(\dfrac{1}{7.9}\)

B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{7.9}\)

B = \(\dfrac{1}{3}\) - \(\dfrac{1}{63}\)

B =  \(\dfrac{20}{63}\)

C = \(\dfrac{5}{2.4.6}\) + \(\dfrac{5}{4.6.8}\) + \(\dfrac{5}{6.8.10}\)

C = \(\dfrac{5}{4}\).(\(\dfrac{4}{2.4.6}\) + \(\dfrac{4}{4.6.8}\) + \(\dfrac{4}{6.8.10}\))

C = \(\dfrac{5}{4}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{4.6}\) + \(\dfrac{1}{4.6}\) - \(\dfrac{1}{6.8}\) + \(\dfrac{1}{6.8}\) - \(\dfrac{1}{8.10}\))

C = \(\dfrac{5}{4}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{8.10}\))

C = \(\dfrac{5}{4}\).( \(\dfrac{1}{8}\) - \(\dfrac{1}{80}\))

C = \(\dfrac{5}{4}\). \(\dfrac{9}{80}\)

C = \(\dfrac{9}{64}\)

Đề sai bạn nhé.

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

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

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

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

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