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Sửa đề . \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{71}{216}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\right)=\frac{71}{216}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{n+2}\right)=\frac{71}{216}\)
\(\Leftrightarrow\frac{1}{n+2}=1-\frac{71}{216}\div\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{n+2}=\frac{37}{108}\)
\(\Leftrightarrow x=\frac{34}{37}\Rightarrow\text{(đề sai) }\)
= 2 x [1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +1/7 -1/9 + .., +1/99 - 1/101
= 2 x [ 1 - 1/101 ]
= 2 x 100/101
= 200/101
t cho mik nha
\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+.........+\(\frac{2}{99.101}\)
=\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+....+\(\frac{1}{99}\)-\(\frac{1}{101}\)
= 1 - \(\frac{1}{101}\)= \(\frac{100}{101}\)
Đặt A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{5}{31}\)
2A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{\left(2x+1\right)}-\frac{1}{2x+3}=\frac{10}{31}\)
2A = \(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
Ta có : \(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
=> 2x + 3 = 93
2x = 90
x = 45
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{5}{36}\)
\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n\left(n+2\right)}\right)=\frac{5}{36}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{18}\)
\(\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)
\(\frac{1}{n+2}=\frac{1}{18}\)
\(\Rightarrow n+2=18\Rightarrow n=16\)
\(\Rightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}=\frac{10}{36}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{18}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)
\(\Rightarrow\frac{n+2-3}{3\left(n+2\right)}=\frac{5}{18}\)
\(\Rightarrow\frac{n-1}{3n+6}=\frac{5}{18}\)
\(\Rightarrow18\left(n-1\right)=5\left(3n+6\right)\)
\(\Rightarrow18n-18=15n+30\)
\(\Rightarrow3n=48\)
\(\Rightarrow n=48:3\)
=>n=16