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Lời giải:
$A=(1+2)+(2^2+2^3)+....+(2^{2020}+2^{2021})$
$=3+2^2(1+2)+....+2^{2020}(1+2)$
$=3+3.2^2+....+3.2^{2020}$
$=3(1+2^2+....+2^{2020})\vdots 3$
Ta có đpcm.
A=(1+2)+(22+23)+...+(210+211)
A=3+22.(1+2)+...+210.(1+2)
A=3+22.3+...+210.3
A=3+(22+...+210)
=>A:cho 3
tick mk nha
A = 1 + 2 + 22 + ... + 211
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{10}+2^{11}\right).\)
\(=3+2^2\left(1+2\right)+2^4\left(1+2\right)+...+2^{10}\left(1+2\right)\)
\(=3\left(1+2^2+2^4+...+2^{10}\right)⋮3\)
A=(1+2)+(2^2+2^3)+...+(2^10+2^11)
= 3+2^2(1+2)+...+2^10(1+2)
=3+2^2.3+...+2^10.3
= 3(1+2^2+...+2^10) chia hết cho 3
=> tổng A chia hết cho 3
\(A=\left(1+2\right)+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)=3+2^2.3+...+2^{10}.3=3\left(1+2^2+...+2^{10}\right)⋮3\)
\(A=1+2+2^2+2^3+...+2^{11}\)
\(A=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{10}+2^{11}\right)\)
\(A=3+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)\)
\(A=3+2^2.3+...+2^{10}.3\)
\(A=3\left(1+2^2+...+2^{10}\right)\)
\(\Rightarrow A⋮3\)
Vậy \(A⋮3\)
!!!
A=(1+2)+(22+23)+...+(210+211)
A=3+22(1+2)+...+210(1+2)
A=3+22.3+...+210.3
A=3(1+22+...+210)chia hết cho 3
=>1+2 +22+23+....+211 chia hết cho 3
\(A=1+2+2^2+2^3+............+2^{11}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{10}+2^{11}\right)\)
\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)\)
\(=\left(1+2\right)\left(1+2^2+...+2^{10}\right)\)
\(=3\cdot\left(1+2^2+...+2^{10}\right)⋮3\)
=>đpcm