Đặt BT trên là A

\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\)

\(2A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\)

\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)

\(2A=\frac{1}{2}-\frac{1}{102}=\frac{50}{102}\Rightarrow A=\frac{25}{102}\)

Đặt A là biểu thức trên ta có : 

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)

\(=\frac{1}{2}\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{102}\right)=\frac{1}{2}.\frac{50}{102}=\frac{25}{102}\)

A = 1×3+3×5+5×7+...+ 97×99+99×101

 6A= 1×3×6+3×5×6+5×7×6+...+97×99×6+99×101×6

6A= 1×3×(5+1)+3×5×(7-1)+5×7×(9-3)+...+97×99×(101-95)+99×101×(103-97)

6A = 1×3×5-1×3+3×5×7-1×3×5+5×7×9-3×5×7+7×9×11-5×7×9+,,,+97×99×101-95×97×99+99×101×103-97×99×101

6A= 1×3+99×101×103

6A= 1029900

A= 171650

171650

A=(1.2)(2.2)+(2.2)(3.2)+...+(50.2)(51.2)

A=1.2.4+2.3.4+...+50.51.4

A=4(1.2+2.3+...+50.51)

M= 1.2+2.3+...+50.51

3M=1.2.3+2.3.(4-1)+...+50.51.(52-49)

=1.2.3+2.3.4-1.2.3+...+50.51.52-49.50.51

= 50.51.52

=132600

=> M=44200

=> A=4M=176800

\(B=2.4+4.6+6.8+...+98.100\)

\(B=2.\left(1.2\right)+2.\left(2.3\right)+2.\left(3.4\right)+...+2.\left(49.50\right)\)

\(B=2\left(1.2+2.3+3.4+....+49.50\right)\)

Đặt:

\(A=1.2+2.3+3.4+...+49.50\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(3A=49.50.51\)

\(A=\dfrac{49.50.51}{3}=41650\)

\(B=2A=41650.2=83300\)

2

\(S1=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)

\(S1=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{51}{102}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\frac{25}{51}\)

\(S1=\frac{25}{102}\)

\(M=\dfrac{3}{2.4}+\dfrac{3}{4.6}+\dfrac{3}{6.8}+.....+\dfrac{3}{100.102}\)

\(M=\dfrac{3.2}{2.4}+\dfrac{3.2}{4.6}+\dfrac{3.2}{6.8}+.....+\dfrac{3.2}{100.102}\)

\(M=3.(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+......+\dfrac{2}{100.102})\)

\(M=3.(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{100}-\dfrac{1}{102})\)

\(M=3.(\dfrac{1}{2}-\dfrac{1}{102})\)

\(M=3.\dfrac{50}{102}\)

\(M=\dfrac{25}{17}\)

Nếu ai mong bn thông cảm!!! Chúc bn hc tốt!

Nếu sai nha m.n

Đặt A = 8.10 + 10.12 + 12.14 + ....... + 98.100

=> 6A = 8.10.12 - 8.10.12 + 10.12.14 - 10.12.14 + ...... + 98.100.102

=> 6A =  98.100.102

=> A = 98.100.102/6

=> A = 166600

c.1.2.3+2.3.4+4.5.6+5.6.7=6+24+120+210

                                      =30+120+210

                                      =150+210

                                      =360

\(M=\dfrac{3}{2.4}+\dfrac{3}{4.6}+\dfrac{3}{6.8}+...+\dfrac{3}{100.102}\)

=> \(2M=\dfrac{3.2}{2.4}+\dfrac{3.2}{4.6}+\dfrac{3.2}{6.8}+...+\dfrac{3.2}{100.102}\)

=> \(2M=3.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{100.102}\right)\)

=> \(2M=3.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{100}-\dfrac{1}{102}\right)\)

=> \(2M=3.\left(\dfrac{1}{2}-\dfrac{1}{102}\right)\)

=> \(2M=3.\dfrac{25}{51}\)

=> \(2M=\dfrac{25}{17}\)

=> \(M=\dfrac{25}{17}:2\)

=> \(M=\dfrac{25}{34}\)

\(A = 1.4 + 2.5 + 3.6 + ...+ 99.102\)

\(A=1.2+1.2+2.3+2.2+3.4+3.2+...+99.100+99.2\)

\(A=(1.2+2.3+3.4+...+99.100)+2.(1+2+3+...+99)\)

\(A=333300+9900\)

\(A=343200\)

\(B = 2.4 + 4.6 + 6.8 + ....+ 98.100 + 100.102\)

\(B=(1.2)(2.2)+(2.2)(3.2)+...+(50.2)(51.2) \)

\(B=4(1.2+2.3+...+50.51) \)

\(M= 1.2+2.3+...+50.51 \)

\(3M=1.2.3+2.3.(4-1)+...+50.51.(52-49) \)

\(=1.2.3+2.3.4-1.2.3+...+50.51.52-49.50.51 \)

\(= 50.51.52\)

\(=132600 \)

\(\Rightarrow\)\(M=44200 \)

\(\Rightarrow\) \(B=4M=176800\)