Ta có : \(\dfrac{1}{2^2}\)<\(\dfrac{1}{1.2}\); \(\dfrac{1}{3^2}\)<\(\dfrac{1}{2.3}\);.....;\(\dfrac{1}{2016^2}\)<\(\dfrac{1}{2015.2016}\)

⇒ A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{2016^2}\)< \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2015.2016}\)

⇒ A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{2016^2}\) < 1 - \(\dfrac{1}{2016}\)= \(\dfrac{2015}{2016}\) (ĐCPCM)

\(vt=1+2015+2015^2+2015^3+2015^4+2015^5+2015^6+2015^7\)

\(=\left(1+2015\right)+\left(2015^2+2015^3\right)+\left(2015^4+2015^5\right)+\left(2015^6+2015^7\right)\)

\(=1\left(1+2015\right)+2015^2\left(1+2015\right)+2015^4\left(1+2015\right)+2015^6\left(1+2015\right)\)

\(=\left(2015+1\right)\left(1+2015^2+2015^4+2015^6\right)\)

\(=2016\left(1+2015^2+2015^4+2015^6\right)\)

\(=2016\left[\left(1+2015^2\right)+\left(2015^4+2015^6\right)\right]\)
\(=2016\left[1\left(1+2015^2\right)+2015^{2014}\left(1+2015^2\right)\right]=vp\left(đpcm\right)\)

\(=2016\left(1+2015^{2014}\right)\left(1+2015^{2012}\right)\)

\(E=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{2015}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 3E=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{2015}{3^{2014}}-\dfrac{2016}{3^{2015}}\\ 3E+E=\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{2015}{3^{2014}}-\dfrac{2016}{3^{2015}}\right)+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{2015}{3^{2015}}-\dfrac{2016}{3^{2016}}\right)\\ 4E=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 4E< 1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}\left(1\right)\)

Gọi \(D=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{2015}}\)

\(3D=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}-\dfrac{1}{3^{2014}}\\ 3D+D=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}-\dfrac{1}{3^{2014}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}\right)\\ 4D=3-\dfrac{1}{3^{2015}}< 3\\ \Rightarrow D< \dfrac{3}{4}\left(2\right)\)

Từ (1) và (2) ta có:

\(4E< \dfrac{3}{4}\\ \Rightarrow E< \dfrac{3}{16}\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2016^2}\)

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)

\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)

\(A< 1-\frac{1}{2016}\)

\(A< \frac{2015}{2016}\left(đpcm\right)\)

\(A=\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{2016.2016}< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2015.2016}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{2015}-\frac{1}{2016}\)

\(=1-\frac{1}{2016}\)

\(=\frac{2015}{2016}\)

\(\Rightarrow A< \frac{2015}{2016}\)

một thửa ruộng hình bình hành có tổng đáy và chiều cao 96m . Cạnh đáy bằng 3/3 chiều cao

A. Tính diện tích thửa ruộng đó.

B.Người ta trồng rau trên thửa ruộng ,cứ 2m vuông thu được 6kg .Tính số rau thu được