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a) \(B\)là tổng các số hạng chia hết cho \(3\)nên chia hết cho \(3\).
b) \(B=3+3^2+...+3^{120}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{119}+3^{120}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{119}\left(1+3\right)\)
\(=4\left(3+3^3+...+3^{119}\right)⋮4\)
c) \(B=3+3^2+...+3^{120}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\)
\(=3\left(1+3+3^2\right)+...+3^{118}\left(1+3+3^2\right)\)
\(=13\left(3+3^4+...+3^{118}\right)⋮13\)
a: \(B=3+3^2+3^3+...+3^{120}\)
\(=3\left(1+3+3^2+...+3^{119}\right)⋮3\)
b: \(B=3+3^2+3^3+3^4+...+3^{2020}\)
\(=3\left(1+3\right)+...+3^{2019}\left(1+3\right)\)
\(=4\cdot\left(3+...+3^{2019}\right)⋮4\)
a: \(A=\left(1+3\right)+...+3^{10}\left(1+3\right)\)
\(=4\left(1+...+3^{10}\right)⋮4\)
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:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
\(B=3+3^2+3^3+...+3^{120}\)
Dễ thấy \(B\)chia hết cho \(3\)do là tổng của các số hạng chia hết cho \(3\).
\(B=3+3^2+3^3+...+3^{120}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{119}+3^{120}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{119}\left(1+3\right)\)
\(=4\left(3+3^3+...+3^{119}\right)⋮4\)
\(B=3+3^2+3^3+...+3^{120}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\)
\(=3\left(1+3+3^2\right)+...+3^{118}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{118}\right)⋮13\)