cái này tính cái gì thế

ko hiểu

M=1−31​+1−151​+1−351​+1−631​+...+1−99991​

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + . . . + \frac{1}{9999} \left.\right)\)

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{99.101} \left.\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)

\(� = 50 - \frac{1}{2} . \left(\right. \frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{5.7} + \frac{2}{7.9} + . . . + \frac{2}{99.101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + . . . + \frac{1}{99} - \frac{1}{101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{101} \left.\right) = 50 - \frac{100}{101} = \frac{5050 - 100}{101} = \frac{4950}{101}\)

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)

\(\frac{1}{2}\)\(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)\)\(=\frac{15}{93}\)

\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{15}{93}:\frac{1}{2}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\rightarrow2x=90\rightarrow x=45\)

\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

= \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)

= \(\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

Bài giải:

\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)

\(=\frac{8}{51}\)

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+.....+\frac{1}{129.15}=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+....+\frac{1}{1935}\)

=\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{43.45}\)

=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{43}-\frac{1}{45}\right)\)

=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{45}\right)\)

=\(\frac{1}{2}.\frac{43}{90}\)

=\(\frac{43}{180}\)

\(S=1:3+1:15+1:35+...+1:9999\)

\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)

\(S=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

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

\(2S=1-\frac{1}{101}\)

\(2S=\frac{100}{101}\)

\(S=\frac{100}{101}:2\)

\(S=\frac{50}{101}\)

 A= 1/1.3+1/3.5+1/5.7+...+1/13.15

2A=2/1.3+2/3.5+2/5.7+...+2/13.15

2A=1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15

2A=1-1/15

2A=14/15

A=7/15

bạn tách mẫu ra mà giải bạn .... naturo làm đúng rồi đó

Ta có: 1/3 ; 1/15 ; 1/35;... 

 <=> 1/1.3 ; 1/3.5 ; 1/5.7

=> chữ số thứ 100 là: 1/199.201

Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{199}-\frac{1}{201}\)

\(=1-\frac{1}{201}=\frac{200}{201}\)

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(B=\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+...+\frac{2}{19×21}\)

\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{19}-\frac{1}{21}\)

\(B=\frac{1}{3}-\frac{1}{21}\)

\(B=\frac{2}{7}\)

A=\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+\(\frac{1}{15}\)+...+\(\frac{1}{66}\) 

A=\(\frac{1}{1\cdot3}\) +\(\frac{1}{2\cdot3}\) +\(\frac{1}{2\cdot5}\)+...+\(\frac{1}{6\cdot11}\)

A=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{5}+...+\frac{1}{6}-\frac{1}{11}\)

A=\(\frac{1}{1}-\frac{1}{11}\)

=>A=\(\frac{10}{11}\)

B=\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\) 

2B=\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{19\cdot21}\)

2B=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

2B=\(\frac{1}{3}-\frac{1}{21}\)

2B=\(\frac{2}{7}\)

B=\(\frac{2}{7}:2\)

=>B=\(\frac{1}{7}\)