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\(S=\dfrac{4}{1.2.3}-\dfrac{1}{1.2.3}+\dfrac{6}{2.3.4}-\dfrac{1}{2.3.4}+...+\dfrac{4018}{2008.2009.2010}-\dfrac{1}{2008.2009.2010}\)
\(=\left(\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2008.2010}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2008.2009.2010}\right)\)
\(=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2008.2010}\right)-\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2008.2009.2010}\right)\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)-\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\)
\(=\left(1-\dfrac{1}{2009}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)-\left(\dfrac{1}{1.2}-\dfrac{1}{2009.2010}\right)\)
\(=1-\dfrac{1}{2009}-\dfrac{1}{2010}+\dfrac{1}{2009.2010}\)
\(=\dfrac{1}{2010}\left(\dfrac{1}{2009}-1\right)-\left(\dfrac{1}{2009}-1\right)\)
\(=\left(\dfrac{1}{2010}-1\right)\left(\dfrac{1}{2009}-1\right)=\dfrac{2009}{2010}.\dfrac{2008}{2009}=\dfrac{1004}{1005}\)
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{637}{1275}\)
\(\Leftrightarrow\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{1}{2}-\dfrac{637}{1275}=\dfrac{1}{2550}\)
\(\Leftrightarrow\left(n+1\right)\left(n+2\right)=2550\)
\(\Leftrightarrow n^2+3n-2548=0\)
\(\Rightarrow n=49\)
\(A=1.2.3...2018\left[\left(1+\dfrac{1}{2018}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2017}\right)+...+\left(\dfrac{1}{1009}+\dfrac{1}{1010}\right)\right]\)
\(A=1.2.3...2018.2019\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\)
\(\dfrac{A}{2019}=1.2.3...2018\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\).
Rõ ràng tích 1 . 2 ... 2018 chia hết cho các tích 1 . 2018; 2 . 2017; ...; 1009 . 1010; do đó \(\dfrac{A}{2019}\) là số tự nhiên.
Vậy A chia hết cho 2019.
Lời giải:
\(A=\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{2011}{1.2.3...2012}\)
\(=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+\frac{4-1}{1.2.3.4}+...+\frac{2012-1}{1.2.3...2012}\)
\(=1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{1.2.3}-\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...2011}-\frac{1}{1.2.3...2012}\)
\(=1-\frac{1}{1.2...2012}< 1\)
Ta có đpcm.
ta có
\(\sqrt{\dfrac{7}{3}}+\sqrt{\dfrac{5}{3}}+1=\dfrac{\sqrt{7}+\sqrt{5}+\sqrt{3}}{\sqrt{3}}\)
tương tự ta có
\(\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{7}{5}}+1=\dfrac{\sqrt{3}+\sqrt{5}+\sqrt{7}}{\sqrt{5}}\)
\(\sqrt{\dfrac{3}{7}}+\sqrt{\dfrac{5}{7}}+1=\dfrac{\sqrt{3}+\sqrt{5}+\sqrt{7}}{\sqrt{7}}\)
\(A=\dfrac{\sqrt{\dfrac{5}{3}}}{\sqrt{\dfrac{7}{3}}+\sqrt{\dfrac{5}{3}}+1}+\dfrac{\sqrt{\dfrac{7}{5}}}{\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{7}{5}}+1}+\dfrac{\sqrt{\dfrac{3}{7}}}{\sqrt{\dfrac{5}{7}}+\sqrt{\dfrac{3}{7}}+1}\)
\(A=\dfrac{\sqrt{5}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}+\dfrac{\sqrt{7}}{\sqrt{7}+\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{7}+\sqrt{5}+\sqrt{3}}=1\)