\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x}=\frac{99}{100}\)
Đặt \(x=n.\left(n+1\right)\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n.\left(n+1\right)}=\frac{99}{100}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{n+1-n}{n.\left(n+1\right)}=\frac{99}{100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{\left(n+1\right)}=\frac{99}{100}\)
\(=1-\frac{1}{\left(n+1\right)}=\frac{99}{100}\)
\(\frac{1}{\left(n+1\right)}=1-\frac{99}{100}=\frac{1}{100}\)
\(\Rightarrow x=\left(100-1\right).100\)
\(=9900\)
Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+......+\frac{1}{x}=\frac{99}{100}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x}=\frac{99}{100}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{x}=\frac{99}{100}\)
Đề kì z