Theo quy luật mà mình nhận thấy thì 20112 phải sửa thành 20122 bạn ạ!
Đặt \(A=\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2011.2013+2}{2012^2}\)
\(\Leftrightarrow A=\frac{2^2+1}{2^2}+\frac{3^2+1}{3^2}+\frac{4^2+1}{4^2}+...+\frac{2012^2+1}{2012^2}\)
\(\Leftrightarrow A=1+\frac{1}{2^2}+1+\frac{1}{3^2}+1+\frac{1}{4^2}+...+1+\frac{1}{2012^2}\)
\(\Leftrightarrow A=\left(1+1+1+...+1\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\right)\)
\(\Leftrightarrow A=2011+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\right)\)
Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
Có: \(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}\)
\(\Leftrightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(\Leftrightarrow B< 1-\frac{1}{2012}\)
\(\Rightarrow A=2011+B< 2011+1-\frac{1}{2012}\)
\(\Rightarrow A< 2012-\frac{1}{2012}< 2013\)
Ta có đpcm
