Question

[Minecraft vn]

\(Cho\)\(A=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2013}\)\(B=\frac{1}{1.2013}+\frac{1}{3.2011}+...+\frac{1}{2011.3}+\frac{1}{2013.1}\)\(Tính\)\(A:B\)

Answer 1

[ST]

\(B=\frac{1}{1.2013}+\frac{1}{3.2011}+...+\frac{1}{3.2011}+\frac{1}{1.2013}\)

\(=\frac{1}{2014}\left(\frac{2014}{1.2013}+\frac{2014}{3.2011}+...+\frac{2014}{1.2013}\right)\)

\(=\frac{1}{2014}\left(\frac{1}{1.2013}+\frac{2013}{1.2013}+\frac{3}{3.2011}+\frac{2011}{3.2011}+...+\frac{2013}{2013.1}+\frac{1}{2013.1}\right)\)

\(=\frac{1}{2014}\left(1+\frac{1}{2013}+\frac{1}{3}+\frac{1}{2011}+...+\frac{1}{2013}+1\right)\)

\(=\frac{2}{2014}\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)\)

\(=\frac{1}{1007}\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2013}}{\frac{1}{1007}\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2013}\right)}=\frac{1}{\frac{1}{1007}}=1007\)

Answer 2

[Hoàng Huyền]

A:B=C