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\(B=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{58}\right)⋮7\)
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
Ta có: A=2+22+23+24+...+299+2100
-> A=2(1+2)+23(1+2)+...+299(1+2)
->A=2.3+23.3+...+299.3
->A=3(2+23+...+299)\(⋮\)3
=> Đpcm
a/ Ta có :
\(A=4+2^2+2^3+...+2^{20}\)
\(\Leftrightarrow2A=8+2^3+2^4+.....+2^{20}+2^{21}\)
\(\Leftrightarrow2A-A=\left(8+2^3+....+2^{21}\right)-\left(4+2^2+...+2^{20}\right)\)
\(\Leftrightarrow A=2^{21}+8-4-2^2\)
\(\Leftrightarrow A=2^{21}\Leftrightarrow\) A là lũy thừa của 2
b/ \(B=2+2^2+2^3+.....+2^{60}\)
\(\Leftrightarrow B=\left(2+2^2\right)+\left(2^3+2^4\right)+.....+\left(2^{59}+2^{60}\right)\)
\(\Leftrightarrow B=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(\Leftrightarrow B=3\left(2+2^3+...+2^{59}\right)⋮3\left(đpcm\right)\)
Các ý sau cũng tương tự !
Thanks bn nha.