Ta có: 5/1x4 + 5/4x7 + ... + 5/100x103

= 5/3 x (1/1 - 1/4 + 1/4 - 1/7 +...+1/100 - 1/103)

= 5/3 x (1 - 1/103)

= 5/3 x 102/103

= 170/103

170/103

ta thấy

1.4=1(2+2)=1.2+1.2=1.2+2

2.5=2(3+2)=2.3+2.2=2.3+4

......................................

100.103=100(101+2)=100.101+100.2=100.101+200

B=1.2+2+2.3+4+3.4+6+...........................+100.101+200

đặt các phép tính nhân là C còn đặt các số tự nhiên là D

tính D trước khoảng cách các số hạng là 2 

co so số hang là :(200-2):2+1=100 số hạng

D= (200+2).100:2=10100

tính C 

ta thấy

1.2=1.2.3/3

2.3=2.3.4/3

................

100.101=100.101.102/3

triệt tiêu các phân số ta có

100.101.102/3-0=343400

vậy B=C+D=343400+10100=353500 

Đặt  \(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+......+\frac{2}{100\cdot103}\)

\(B=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{2}{3}\cdot\left(1-\frac{1}{103}\right)\)

\(B=\frac{2}{3}\cdot\frac{102}{103}\)

\(\Rightarrow B=\frac{68}{103}\)

Đặt \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)

\(A=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(A=\frac{2}{3}\left(1-\frac{1}{103}\right)\)

\(A=\frac{2}{3}\cdot\frac{102}{103}\)

\(A=\frac{68}{103}\)

\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)

\(=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{2}{3}\cdot\left(1-\frac{1}{103}\right)\)

\(=\frac{2}{3}\cdot\frac{102}{103}=\frac{68}{103}\)

\(F=\dfrac{1}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{100\cdot103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{102}{103}=\dfrac{34}{103}\)

Giờ anh đang bận hồi nữa anh giúp cho nha

\(F=\dfrac{1}{x}\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}+\dfrac{1}{100.103}\right)\)

\(3F=\dfrac{1}{x}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}+\dfrac{3}{100.103}\right)\)

\(F=\dfrac{\dfrac{1}{x}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)}{3}=\dfrac{\dfrac{1}{x}.\dfrac{100}{309}}{3}=\dfrac{\dfrac{100x}{309}}{3}=\dfrac{100x}{927}\)

\(F=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{204}{309}\)

\(a,\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{43.45}=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{43.45}\right)=\frac{5}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{45}\right)=\frac{5}{3}.\frac{44}{45}=\frac{44}{27}\)

S=1/1-1/4+1/4+1/7-1/7+1/10+...+1/100-1/103

S=1/1-1/103

S=102/103

Vì 102/103<1 nên S<1

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{100\cdot103}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)

\(S=1-\frac{1}{103}\)

\(S=\frac{102}{103}< 1\)

\(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+.......+\frac{3}{100x103}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}....+\frac{1}{100}-\frac{1}{103}\)

\(=\frac{1}{1}-\frac{1}{103}\)

=\(\frac{102}{103}\)