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Xét \(4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2014}{4^{2013}}\)
=> \(3S=4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2014}{4^{2013}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+...+\dfrac{2014}{4^{2014}}\right)\)
=> \(3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}-\dfrac{2014}{4^{2014}}< 1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)
Đặt \(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)
=> \(4A=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\)
=> \(3A=4A-A=\left(4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\right)\)
=> \(3A=4-\dfrac{1}{4^{2013}}< 4\)
=> \(A< \dfrac{4}{3}\)
=> \(3S< \dfrac{4}{3}\)
=> \(S< \dfrac{4}{9}< \dfrac{1}{2}\)
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}\)
\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2014}{4^{2014}}\right)\)
\(3S=1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+......+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.....+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
đặt \(A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2023}}\)
\(4A-A=4+1+\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{2022}}-\left(1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2023}}\right)\)
\(3A=4-\frac{1}{4^{2023}}\)
\(A=\frac{4}{3}-\frac{1}{3.4^{2023}}\)
\(\Rightarrow3S=\frac{4}{3}-\frac{1}{3.4^{2023}}-\frac{2014}{4^{2024}}\)
\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}\)
do \(\frac{4}{9}< \frac{4}{8}=\frac{1}{2}\)
\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}< \frac{4}{8}=\frac{1}{2}\left(đpcm\right)\)
\(A=2014.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2013}\right)\)
\(A=2014.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1007.2013}\right)\)
\(A=2.2014.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2014}\right)\)
\(A=2.2014.\frac{2013}{2014}\)
\(A=\frac{2.2014.2013}{2014}\)
\(A=2.2013\)
\(A=4026\)
\(ĐặtA=\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2014}{2^{2013}}+\frac{2015}{2^{2014}}\)
\(2A=\frac{3}{2}+\frac{4}{2^2}+...+\frac{2014}{2^{2012}}+\frac{2015}{2^{2013}}\)
\(2A-A=\left(\frac{3}{2}+\frac{4}{2^2}+...+\frac{2014}{2^{2012}}+\frac{2015}{2^{2013}}\right)-\left(\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2014}{2^{2013}}+\frac{2015}{2^{2014}}\right)\)
\(A=\frac{3}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}+\frac{1}{2^{2013}}-\frac{2015}{2^{2014}}\)
\(2A=3+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}+\frac{1}{2^{2012}}-\frac{2015}{2^{2013}}\)
\(2A-A=\left(3+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}+\frac{1}{2^{2012}}-\frac{2015}{2^{2013}}\right)-\left(\frac{3}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}+\frac{1}{2^{2013}}-\frac{2015}{2^{2014}}\right)\)
\(A=3+\frac{1}{2}-\frac{2015}{2^{2013}}-\frac{3}{2}-\frac{1}{2^{2013}}+\frac{2015}{2^{2014}}\)
\(A=2-\frac{2015}{2^{2013}}-\frac{1}{2^{2013}}+\frac{2015}{2^{2014}}\)
\(A=2-\frac{4030}{2^{2014}}-\frac{2}{2^{2014}}+\frac{2015}{2^{2014}}\)
\(A=2-\frac{4032}{2^{2014}}+\frac{2015}{2^{2014}}\)
\(A=2-\frac{2017}{2^{2014}}< 2\)
=> đpcm
Bài này dễ thôi mà nhưng mình chỉ gợi ý thôi nhé! Bạn phải đổi phần mẫu số ra đã nhé ! *CỐ LÊN*
có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1
=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)
vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015
x=2015
gọi dãy số trên là A
ta có A<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
A<1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
A<1-\(\frac{1}{2014}\)=\(\frac{2013}{2014}\)
Vậy A < \(\frac{2013}{2014}\)
ko biết