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B=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.....+\frac{1}{3^{2012}}+\frac{1}{3^{2013}}\)
3B=\(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2011}}+\frac{1}{3^{2012}}\)
3B-B=\(\left(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2011}}+\frac{1}{3^{2012}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2012}}+\frac{1}{3^{2013}}\right)\)
2B=\(1-\frac{1}{3^{2013}}\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\)
\(3B=\frac{1}{3}.3+\frac{1}{3^2}.3+\frac{1}{3^3}.3+...+\frac{1}{3^{2013}}.3\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2012}}\)
\(3B-B=2B=\)
3B= \(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2012}}\)
B= \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2012}}+\frac{1}{3^{2013}}\)
2B= 1 + 0 + 0 + 0 +.......+ 0 - \(\frac{1}{3^{2013}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2013}}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2013}}\)
\(\Rightarrow B< \frac{1}{2}\)
Vậy \(B< \frac{1}{2}\).
tại vì có cộng bao nhiêu số thì khi rút gọn cung ko thể lớn hơn 4/9 vì 4/9 còn có thể là 40000000/90000000 nên là ko thể
Ta có :\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}=\frac{1}{2}-\frac{1}{2017}=\frac{2015}{4034}< \frac{1}{2}< \frac{4}{9}\)(đpcm)
I don't now
or no I don't
..................
sorry
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2001^2}+\frac{1}{2002^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2000.2001}+\frac{1}{2001.2002}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}\)
\(\Rightarrow A< 1-\frac{1}{2002}=\frac{2001}{2002}\left(đpcm\right)\)
Ta có: \(B=\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}\)
\(B=1+\left(\frac{1}{2016}+1\right)+\left(\frac{2}{2015}+1\right)+\left(\frac{3}{2014}+1\right)+...+\left(\frac{2015}{2}+1\right)\)
\(B=\frac{2017}{2017}+\frac{2017}{2016}+\frac{2017}{2015}+\frac{2017}{2014}+...+\frac{2017}{2}\)
\(B=2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)}\)
\(\Rightarrow\frac{A}{B}=\frac{1}{2017}.\)
Chúc bạn học tốt!
Này Vũ Minh Tuấn, mk cũng có 1 bài cũng gần giống như thế này nhưng khác 1 tí cậu giải giúp mk vs
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2015^2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2015}\)
\(\Rightarrow A< \approx0,75\)
Vậy.....