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

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

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

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

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

Sau bạn tính tiếp là OK rồi 

Hình như đề bài nhầm thì phải: 61/63 chứ?

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

\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+...+\dfrac{9997}{9999}\)

\(=1-\dfrac{2}{3}+1-\dfrac{2}{15}+1-\dfrac{2}{35}+...+1-\dfrac{2}{9999}\)

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

\(=50-1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

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

\(=\dfrac{4950}{101}\)

\(\frac{1}{3}+\frac{13}{15}+\frac{33}{35}+...+\frac{9997}{9999}=1-\frac{2}{3}+1-\frac{2}{15}+1-\frac{2}{35}+...+1-\frac{2}{9999}\)

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

\(=50-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

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

thank you bạn nhé mình sẽ k cho bạn

nhưng mà sao bạn biết là có 50 số 1

\(\frac{1}{3}+\frac{13}{15}+\frac{33}{35}+\frac{61}{63}+\frac{97}{99}\)\(+\frac{141}{143}\)

\(=\left(1-\frac{2}{3}\right)+\left(1-\frac{2}{15}\right)\)\(+\left(1-\frac{2}{35}\right)+\left(1-\frac{2}{63}\right)\)\(+\left(1-\frac{2}{99}\right)+\left(1-\frac{2}{143}\right)\)

\(=\left(1+1+1+1+1+1\right)-\)\(\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)

\(=6-\)\(\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+\frac{2}{11\times13}\right)\)

\(=6-\)\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\left(1-\frac{1}{13}\right)\)

\(=6-\frac{12}{13}\)

\(=\frac{66}{13}\)

\(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{9603}+\dfrac{2}{9999}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\cdot\dfrac{96}{505}=\dfrac{150}{101}\)