=> 4E = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 99.100.101.4

=> 4E = 1.2.3.( 4 - 0 ) + 2.3.4.( 5 - 1 ) + 3.4.5.( 6 - 2 ) + .... + 99.100.101.( 102 - 98 )

=> 4E = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 99.100.101.102 - 98.99.100.101

=> 4E = ( 1.2.3.4 - 1.2.3.4 ) + ( 2.3.4.5 - 2.3.4.5 ) +  + ... + ( 98.99.100.101 - 98.99.100.101 ) + 99.100.101.102

=> 4E = 99.100.101.102

=> E = ( 99.100.101.102 ) : 4

Ô tô đi với vận tốc 50km/giờ vì :

         100 : 2 = 50

                   đs : 50

Ta có: A = 1.2.3+3.4.5+5.6.7+...+99.100.101

A = 1.3 (5-3) + 3.5 (7-3) + 5.7 (9-3) + ............ + 99.101 (103 - 3)

A = (1.3.5 + 3.5.7 + 5.7.9 + .......... + 99.101.103) - (1.3.3 + 3.5.3 + ....... + 99.101.3)

A = (15+99.101.103.105) : 8 - 3.(1.3 + 3.5 +5.7 + ...... + 99.101)

A = 13517400 - 3.171650

A = 13002450

1.2.3.4+2.3.4.5+3.4.5.6+...+97.98.99.100

4S=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100). 4

4S=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4S=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...98.99.100.101-97.98.99.100

4S=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98+99.100+101

4S=98.99.100.101

Vậy S = 98.99.100.101/4 = 24497550

\(A=1.2.3+3.4.5+5.6.7+...+99.100.+101\)

\(A=1.3\left(5-3\right)+3.5\left(7-3\right)+5.7\left(9-3\right)+...+99.100\left(103-3\right)\)

\(=\left(1.3.5+3.5.7+5.7.9+99.101.103\right)-\left(1.3.3+3.5.3+99.101.3\right)\)

\(=\left(15+99.101.103.105\right):8-3.\left(1.3+3.5+5.7+99.101\right)\)

\(=13517400-3.171650\)

\(=13002450\)

cau còn lại bạn trả lời được mình sẽ tích cho bạn luôn

Đúng ko 

Rút gọn mỗi số hãng của số ta được :

\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

     \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

     \(=1-\frac{1}{101}=\frac{100}{101}\)

Vậy C = 100/101

\(C=\frac{4}{1.2.3}+\frac{8}{3.4.5}+\frac{12}{5.6.7}+...+\frac{200}{99.100.101}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{101}{101}-\frac{1}{101}\)

\(=\frac{100}{101}\)

\(C=\frac{4}{1.2.3}+\frac{8}{3.4.5}+\frac{12}{5.6.7}+.....+\frac{200}{99.100.101}\)

\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)

\(C=\frac{1}{2}\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+.....+\frac{2}{99}-\frac{2}{101}\right)\)

\(C=\frac{1}{2}\left(2-\frac{2}{101}\right)\)

\(C=\frac{1}{2}\left(\frac{202}{101}-\frac{2}{101}\right)\)

\(C=\frac{1}{2}.\frac{200}{101}\)

\(C=\frac{200}{202}=\frac{100}{101}\)