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

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

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

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

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

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

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

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

=1/6-1/406

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

\(\Leftrightarrow I=\frac{1}{4}\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{402}-\frac{1}{406}\right)\)

\(\Leftrightarrow I=\frac{1}{4}\left(\frac{1}{6}-\frac{1}{406}\right)\)

\(\Leftrightarrow I=\frac{1}{4}\cdot\frac{100}{609}\)

\(\Leftrightarrow I=\frac{25}{609}\)

4I = 4/6.10 + 4/10.14 + ......+ 4/402.406   À mình nói thêm tử số sẽ dựa vào khoảng cách giữa 2 mẫu số

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

    = 1/6 - 1/406 = 100/609

I   =100/609 : 4 = 25/609 

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 :

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