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a) Giải
Ta có: \(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}+\dfrac{1}{2^{2013}}\)
\(\Rightarrow2S=\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+\dfrac{2}{2^{2012}}+\dfrac{2}{2^{2013}}\)
\(2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)
\(\Rightarrow2S-S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2012}}-\dfrac{1}{2^{2013}}\)
\(\Rightarrow S=1-\dfrac{1}{2^{2013}}\)
\(\Rightarrow S=\dfrac{2^{2013}-1}{2^{2013}}\)
b) Giải
Từ \(A=\dfrac{2011^{2012}+1}{2011^{2013}+1}\)
\(\Rightarrow2011A=\dfrac{2011^{2013}+20111}{2011^{2013}+1}=\dfrac{2011^{2013}+1+2010}{2011^{2013}+1}=1+\dfrac{2010}{2011^{2013}+1}\)
Từ \(B=\dfrac{2011^{2013}+1}{2011^{2014}+1}\)
\(\Rightarrow2011B=\dfrac{2011^{2014}+2011}{2011^{2014}+1}=\dfrac{2011^{2014}+1+2010}{2011^{2014}+1}=1+\dfrac{2010}{2011^{2014}+1}\)
Vì 20112013 + 1 < 20112014 + 1 và 2010 > 0
\(\Rightarrow\dfrac{2010}{2011^{2013}+1}>\dfrac{2010}{2011^{2014}+1}\)
\(\Rightarrow2011A>2011B\)
\(\Rightarrow A>B\)
Vậy A > B.
\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2014}}\\ 3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\\ 3A-A=\left(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2014}}\right)\\ 2A=3-\dfrac{1}{3^{2014}}\\ A=\left(3-\dfrac{1}{3^{2014}}\right):2\\ A=3:2-\dfrac{1}{3^{2014}}:2\\ A=\dfrac{3}{2}-\dfrac{1}{3^{2014}\cdot2}< \dfrac{3}{2}\)
Vậy \(A< \dfrac{3}{2}\)
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\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)....\left(\dfrac{1}{100^2}-1\right)\\ =-\dfrac{3}{4}.-\dfrac{8}{9}...-\dfrac{9999}{10000}\\ =-\left(\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\right)\\ =-\left(\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}....\dfrac{99.101}{100.100}\right)=-\left(\dfrac{1.2....99}{2.3....100}.\dfrac{3.4....101}{2.3....100}\right)\\ =-\left(\dfrac{1}{100}.\dfrac{101}{2}\right)\\ =-\dfrac{101}{200}< \dfrac{-100}{200}=-\dfrac{1}{2}\\ \Rightarrow A< \dfrac{-1}{2}\)
=1/1.2+1/1.2.3+.............1/1.2.3.4. ......... .2019
=1-1/2+1/2-1/3+1/3-.............-2019
=1-1/2019
=2018/2019
Vay 1/1.2+1/1.2.3+.............1/1.2.3.4. ......... .2019>1/2
\(3S=1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\)
=>2S=1-1/3^100
=>S=1/2-1/2*3^100<1/2