Đây là bài toán tìm tổng dãy số có quy luật.
Để ý thấy rằng \(\frac{1}{n\left(n+2\right)}=\frac{1}{2}.\frac{2}{n\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)\)
Vậy thì \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{n+2}\right)=\frac{5}{36}\Rightarrow\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)
\(\Rightarrow\frac{1}{n+2}=\frac{1}{18}\Rightarrow n=16.\)
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{n\left(n+2\right)}=\frac{5}{36}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{36}\)
\(\frac{1}{3}-\frac{1}{n+2}=\frac{5}{36}\)
\(\frac{12}{36}-\frac{1}{n+2}=\frac{5}{36}\)
\(\frac{1}{n+2}=\frac{7}{36}\)
\(\Rightarrow\frac{7}{7\left(n+2\right)}=\frac{7}{36}\)
\(7\left(n+2\right)=36\)
n + 2 = 36/7
n = 36/7 - 2
( Tự tính KQ nha )
1/3.5+1/5.7+1/7.9+....+1/n(n+2)=5/36
=1(1/3.5+1/5.7+1/7.9+....+1/n(n+2)
=1/2(2/3.5+2/5.7+2/7.9+....+2/n(n+2)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/n-1/n+2
=1/2(1/3+-1/5+1/5+-1/7+1/7+-1/9+...+1/n+-1/n+2
=1/2(1/3+0+0......+-1/n+2)=5/36
=1/2(1/3+1/n+2)=5/36
=(1/3+1/n+2=5/36:1/2
=1/3+1/n+2=5/18
=1/n+2=5/18-1/3
1/n+2=-1/18
=1/n+2=1/-18
=1/n=1/-18-2
=1/n=1/-20
n=-20
=
