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Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\)(Tự chứng minh)
Nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2015}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015}\)
Gọi \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015}\)
Ta có: \(A=\frac{1}{1}-\frac{1}{2015}+\frac{1}{2015}=1\)
Do đó \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2015}<1\)
\(\frac{1}{2^2}>\frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}>\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(....\)
\(\frac{1}{2015^2}>\frac{1}{2014.2015}=\frac{1}{2014}-\frac{1}{2015}\)
nên \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2015^2}>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2004}-\frac{1}{2005}\)
\(=1-\frac{1}{2005}\)
vì \(1-\frac{1}{2005}< 1\)
=> ĐPCM
xét mẫu(chỗ 1/2014 sửa lại thành 2/2014)
=(1/2015+1)+(2/2014+1)+...+(2013/3+1)+(2014/2+1)+(2015/1-2014)
=2016/2015+2016/2014+...+2016/3+2016/2+1
=2016.(1/2016+1/2015+...+1/4+1/3+1/2)
=> A= 1/2016
mún dễ hỉu hơn hãy gửi tin nhắn cho mik
Mẫu số = \(\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)
= \(1+1+1+...+1\) ( có tổng cộng 2015 số 1) \(+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)
= \(\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)\)
= \(\left(\frac{2}{2}+\frac{2014}{2}\right)+\left(\frac{3}{3}+\frac{2013}{3}\right)+...+\left(\frac{2015}{2015}+\frac{1}{2015}\right)\)
= \(\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}\)
= \(2016.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}\right)\)
Tử số= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}\)
Lấy tử số chia cho mẫu số:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}}{2016.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}\right)}\)
Đơn giản mẫu và tử.
\(A=\frac{1}{2016}\)
\(TA-CO':\)
\(A=\frac{4+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}{7+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}\)
\(A=\frac{4\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}{7\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(A=\frac{4}{7}\)
\(B=\frac{1+2+...+2^{2013}}{2^{2015}-2}\)
ĐẶT \(C=1+2+...+2^{2013}\)
\(\Rightarrow2C=2+2^2+...+2^{2014}\)
\(\Rightarrow2C-C=\left(2+2^2+...+2^{2014}\right)-\left(1+2+...+2^{2013}\right)\)
\(\Rightarrow C=2^{2014}-2\)
\(\Rightarrow B=\frac{2^{2014}-1}{2^{2015}-2}\)
\(B=\frac{2^{2014}-1}{2\left(2^{2014}-1\right)}\)
\(B=\frac{1}{2}\)
\(\Rightarrow A-B=\frac{3}{7}-\frac{1}{2}=\frac{6}{14}-\frac{7}{14}\)
\(A-B=\frac{6-7}{14}=\frac{-1}{14}\)
VẬY, \(A-B=\frac{-1}{14}\)