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\(\frac{1}{1.\left(2n-1\right)}+\frac{1}{3.\left(2n-3\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}\)
\(=\frac{1}{2n}\left[\frac{2n-1+1}{1\left(2n-1\right)}+\frac{2n-3+3}{3\left(2n-3\right)}+...+\frac{3+2n-3}{\left(2n-3\right).3}+\frac{1+2n-1}{\left(2n-1\right).1}\right]\)
\(=\frac{1}{2n}\left(1+\frac{1}{2n-1}+\frac{1}{3}+\frac{1}{2n-3}+...+\frac{1}{2n-3}+\frac{1}{3}+\frac{1}{2n-1}+1\right)\)
\(=\frac{1}{n}\left(1+\frac{1}{3}+...+\frac{1}{2n-3}+\frac{1}{2n-1}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{1}{n}\).
Lời giải:
\(A=\frac{1}{1(2n-1)}+\frac{1}{3(2n-3)}+...+\frac{1}{(2n-3).3}+\frac{1}{(2n-1).1}\)
\(2nA=\frac{1+(2n-1)}{1(2n-1)}+\frac{3+(2n-3)}{3(2n-3)}+....+\frac{(2n-3)+3}{(2n-3).3}+\frac{(2n-1)+1}{(2n-1).1}\)
\(2nA=\frac{1}{2n-1}+1+\frac{1}{2n-3}+\frac{1}{3}+...+\frac{1}{3}+\frac{1}{2n-3}+1+\frac{1}{2n-1}\)
\(=\left(\frac{1}{2n-1}+\frac{1}{2n-3}+...+\frac{1}{3}+1\right)+\left(1+\frac{1}{3}+...+\frac{1}{2n-3}+\frac{1}{2n-1}\right)\)
\(=2\left(1+\frac{1}{3}+...+\frac{1}{2n-1}\right)\)
\(\Rightarrow A=\frac{1}{n}\left(1+\frac{1}{3}+...+\frac{1}{2n-1}\right)\)
\(B=\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}....\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)
\(=\frac{1^2}{\left(2-1\right)\left(2+1\right)}.\frac{3^2}{\left(4-1\right)\left(4+1\right)}...\frac{\left(2n+1\right)^2}{\left(2n+2-1\right)\left(2n+2+1\right)}\)
\(=\frac{1^2}{1.3}.\frac{3^2}{3.5}...\frac{\left(2n+1\right)^2}{\left(2n+1\right)\left(2n+3\right)}\)
\(=\frac{2n+1}{2n+3}\)
P/S: tham khảo nha, mk ko chắc là đúng
