Ta có \(6B=1\times3\times6+3\times5\times6+...+97\times99\times6\)

\(=1\times3\times\left(5+1\right)+3\times5\times\left(7-1\right)+5\times7\times\left(9-3\right)+...+97\times99\times\left(101-95\right)\)

\(=1\times3\times5+1.3+3\times5\times7-3\times5\times1+...-97\times99\times95\)

\(=97\times99\times101+3\)

\(\Rightarrow B=\frac{97\times99\times101+3}{6}=161651\)

6B=1x3x6+3x5x6+5x7x6+.....+97x99x6

6B=1x3x(5+1)+3x5x(7-1)+....+97x99x(102-95)

6B=1x3x5+1x3+3x5x7-3x5+....+97x99x101-95x97x99

6B=1x3x97x99x101

6B=969906

=>B=161651

6B = 1x3x6 + 3x5x6 + 5x7x6 +...+ 99x101x6

     = 1x3(5+1)+3x5(7-1)+...+97x99(101-95)

     = 1x3x5+1x3+3x5x(7-1)+...-97x99x95

     =97x99x101+3       rồi tự làm.....................  >.<

B=3×53​+5×73​+7×93​+...+48×503​ 

\(B = \frac{3}{2} \times \left(\right. \frac{2}{3 \times 5} + \frac{2}{5 \times 7} + \frac{2}{7 \times 9} + . . . + \frac{2}{48 \times 50} \left.\right)\) 

\(B = \frac{3}{2} \times \left(\right. \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + . . . + \frac{1}{48} - \frac{1}{50} \left.\right)\) 

\(B = \frac{3}{2} \times \left(\right. \frac{1}{3} - \frac{1}{50} \left.\right)\)

\(B = \frac{3}{2} \times \frac{47}{150}\) 

\(B = \frac{47}{100}\) 

= 1-1/3 + 1/3-1/5+.......+1/97-1/99

=  1 - 1/99

= 98/99

sao lại là 1- 1/3 + 1/3 -1/5 + ...... 1/97 - 1/99 hả bạn :|

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{95.97}\)\(+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{98}{99}=\dfrac{1}{33}\cdot49=\dfrac{49}{33}\)

sửa đề câu a  và câu b  nhá  , mik nghĩ đề như này :

  \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

 \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

= \(\frac{1}{1}-\frac{1}{215}\)

\(=\frac{214}{215}\)

b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)

    \(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)

\(A\cdot2=\frac{214}{215}\)

\(A=\frac{214}{215}:2\)

\(A=\frac{107}{215}\)

@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều

trả lời hiền thương đề bài của bạn ấy là đúm gòi nha

A = 2/3 x 5+ 2/5 x 7 + 2/7 x 9 + ... + 2/97 x 99

A = 2/3 - 2/99

A = 64/99

A=2/3x5 + 2/5x7 + 2/7x9 + ......+ 2/97x99 = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99 
A= 1/3 - 1/99 = 96/3.99 = 32/99 

        Đ/S:tích

A = 2/3 x 5 + 2/ 5 x 7 + 2/7 x 9 + ... + 2/97 x 99

A = 1/3 - 1/99

A = 32/99