B=1x3+3x5+5x7+7x9+...+95x97+97x99

= 1.(1+2)+3.(3+2)+5.(5+2)+....+95.(95+2)+97.(97+2)

= 1²+1.2+3²+3.2 +5²+5.2+...+95²+95.2+ 97²+97.2

= (1²+3² +5²+...+95²+ 97²)+(1.2+3.2 +5.2+...+95.2+97.2)

= (1²+3² +5²+...+95²+ 97²)+ 2.(1+3 +5+...+95+97)

Đặt : A = 1²+3² +5²+...+95²+ 97² 

C =1+3 +5+...+95+97  

    tính A và C (tìm câu hỏi tương tự hình như anh thấy họ làm rồi đấy) sau đó thay vào tính B 

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

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

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

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

b)

S2=6/2x5+6/5x8+6/8x11+...+6/29x32

=2.(3/2.5+3/5.8+...+3/29.32)

=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2.(1/2-1/32)

=2.15/32

=15/16

a)

Ta có:

S1=2/3x5+2/5x7+2/7x9+...+2/97x99

=1/3-1/5+1/5-1/7+...+1/97-1/99

=1/3-1/99

=32/99

M=\(\frac{2}{3\times5}+\frac{2}{5\times7}+.............+\frac{2}{97\times99}\)

=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..........+\frac{1}{97}-\frac{1}{99}\)

=\(\frac{1}{3}-\frac{1}{99}\)

=\(\frac{32}{99}\)

\(\Leftrightarrow M=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{97}-\frac{2}{99}\)

\(\Rightarrow M=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow M=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(\Rightarrow M=2\times\frac{32}{99}\)

\(\Rightarrow M=\frac{64}{99}\)