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\(2S=2-2^2+2^3-2^4+...+2^{2013}\)
\(\Leftrightarrow3S=2^{2013}+1\)
\(\Leftrightarrow3S-2^{2013}=1\)
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=1−3+5−7+...+2001−2003+2005S=1−3+5−7+...+2001−2003+2005
=(1−3)+(5−7)+...+(2001−2003)+2005=(1−3)+(5−7)+...+(2001−2003)+2005(Có 1002 cặp)
=(−2).1002+2005=(−2).1002+2005
=−2004+2005=−2004+2005
=1
https://hoc247.net/hoi-dap/toan-6/chung-minh-s-1-2-2-2-2-3-2-4-2-5-2-6-2-7-chia-het-cho-3-faq250754.html
S= \(1+2+2^2+...+2^7\)
2S= \(2\cdot\left(2+2^2+...+2^7\right)\)
2S= \(2^1+2^2+...2^8\)
1S= 2S - S = \(\left(2^1+2^2+...2^8\right)-\left(1+2+2^2+...+2^7\right)\)
1S= \(2^1+2^2+...+2^8-1-2-2^2-...-2^7\)
1S= \(2^8-1\)
1S= \(256-1\)
1S= 255
=> 1S chia hết cho 3
Mà 1S= S
=> S chia hết cho 3
Vậy S chia hết cho 3
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