3/15+3/35+3/63+...+3/9603

=3/3.5+3/5.7+3/7.9+...+3/97.99

=3/2.(2/3.5+2/5.7+2/7.9+...+2/97.99)

=3/2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)=3/2.(1/3-1/99)

=3/2.32/99

=16/33

\(\frac{16}{33}\)

Ta gọi A= 3/15+3/35+3/63+...+3/9603

A=3/3.5+3/5.7+3/7.9+...+3/97.99

2.A=3.2/3.5+3.2/5.7+...+3.2/97.99

2.A=3/3-3/5+3/5-3/7+...+3/97-3/99

2.A=3/3-3/99

2.A=32/33

  A=32/33:2

  A=16/33

=3*(1 /(3*5)+1 /(5*7)+...+1/ (97*99))

=3*1/2(1/3-1/5+1/5-1/7+...+1/97-1/99)

=3/2*(1/3-1/99)

=16/33

1/3+13/15+33/35+31/63+.....................+9601/9603+9997/9999

\(=1-\frac{2}{3}+1-\frac{2}{15}+...+1-\frac{2}{9999}\)

\(=\left(1+1+1+1+...+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\right)\)

\(=50-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

\(=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{4950}{101}\)

HTDT

có: 2/3=1-1/3

2/15=1/3-1/5

...

=> <1

\(\frac{1}{3}+\frac{13}{15}+...+\frac{9997}{9999}\)

\(=1-\frac{2}{3}+1-\frac{2}{15}+...+1-\frac{2}{9999}\)

\(=\left(1+1+...+1\right)-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

\(=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=50-\left(1-\frac{1}{101}\right)\)

Sau bạn tính tiếp là OK rồi 

Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)

\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

\(A=\frac{5}{39}\)

Câu còn lại cx dựa như vậy nhé bn ! 

Chúc bn hc tốt <3

câu c hình như sai đề hả bn