Ta có công thức :
\(\frac{1}{k\left(k+1\right)}=\frac{\left(k+1\right)-k}{k\left(k+1\right)}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}=\frac{n-1}{n}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\)
\(A=1-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{n-1}+\frac{1}{n+1}-\frac{1}{n}\)
\(=1-0-0-0-...-0-\frac{1}{n}\)
\(=\frac{n-1}{n}\)
Ta có : \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{\left(n-1\right)n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}=\frac{n-1}{n}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{\left(n-1\right)\cdot n}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\)
\(A=\frac{1}{1}-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}\)
A=đã cho
Áp dụng công thức \(\frac{n}{x\cdot\left(x+n\right)}=\frac{1}{x}-\frac{1}{x+n}\)
=>A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)
Xin lỗi viết nhầm đoạn cuối,phải là \(+\frac{1}{n-1}-\frac{1}{n}\)
=>\(A=1-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}\)
Vậy....
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow A=1-\frac{1}{n}=\frac{n-1}{n}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}\)
