Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
B = \(1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\right)\)
Xét \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}=1-\frac{1}{2016}=\frac{2015}{2016}\)
Do đó B > 1 - \(\frac{2015}{2016}=\frac{1}{2016}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2001^2}+\frac{1}{2002^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2000.2001}+\frac{1}{2001.2002}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}\)
\(\Rightarrow A< 1-\frac{1}{2002}=\frac{2001}{2002}\left(đpcm\right)\)
4A = 1 +1/2^2+1/2^4+....+1/2^98
3A = 4A-A = (1+1/2^2+1/2^4+....+1/2^98) - (1/2^2+1/2^4+....+1/2^100) = 1 - 1/2^100 < 1
=> A < 1/3 ( ĐPCM )
k mk nha
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{2016}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2016}}\)
\(2A=2+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\)
\(2A-A=2+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^{2015}}-\frac{1}{2}-\frac{1}{2^2}-..-\frac{1}{2^{2016}}\)
\(A=2-\frac{1}{2^{2016}}\)
\(A=\frac{2^{2009}}{2^{2010}}-\frac{1}{2^{2010}}=\frac{2^{2009}-1}{2^{2010}}\)
\(\Rightarrow A< 1\)