Question

Tìm số tự nhiên n biết : \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{n.\left(n+2\right)}\)\(=\frac{5}{36}\)

Giúp mình một tí nha !

Answer 1

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{5}{36}\)

\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n\left(n+2\right)}\right)=\frac{5}{36}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{18}\)

\(\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)

\(\frac{1}{n+2}=\frac{1}{18}\)

\(\Rightarrow n+2=18\Rightarrow n=16\)

Answer 2

\(\Rightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}=\frac{10}{36}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{18}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)

\(\Rightarrow\frac{n+2-3}{3\left(n+2\right)}=\frac{5}{18}\)

\(\Rightarrow\frac{n-1}{3n+6}=\frac{5}{18}\)

\(\Rightarrow18\left(n-1\right)=5\left(3n+6\right)\)

\(\Rightarrow18n-18=15n+30\)

\(\Rightarrow3n=48\)

\(\Rightarrow n=48:3\)

=>n=16