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S = 3+3^2 + 3^3 +...+ 3^2016
= (3+3^2+3^3) +...+(3^2014+3^2015+3^2016)
=3(1+3+3^2) +.....+3^2014(1+3+32)
=13 ( 3+...+3^2014 ) chia hết cho 13
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(S=1+3^1+3^2+3^3+...+3^{2017}+3^{2018}\)
\(=\left(1+3^1+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{2016}+3^{2017}+3^{2018}\right)\)
\(=13+3^3\cdot13+...+3^{2016}\cdot13\)
\(=13\cdot\left(1+3^3+...+3^{2016}\right)⋮13\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=1+(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2014}+3^{2015}+3^{2016})$
$=1+3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2014}(1+3+3^2)$
$=1+3.13+3^4.13+....+3^{2014}.13$
$=1+13(3+3^4+...+3^{2014})$
$\Rightarrow A-1\vdots 13(1)$
Mặt khác:
$A=1+(3+3^2+3^3+3^4)+....+(3^{2013}+3^{2014}+3^{2015}+3^{2016})$
$=1+3(1+3+3^2+3^3)+....+3^{2013}(1+3+3^2+3^3)$
$=1+(3+...+3^{2013})(1+3+3^2+3^3)$
$=1+40(3+....+3^{2013})$
$\Rightarrow A-1\vdots 5(2)$
Từ $(1); (2)$ mà $(5,13)=1$ nên $A-1\vdots (5.13)$ hay $A-1\vdots 65$
$\Rightarrow A$ chia $65$ dư $1$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:3x3x3x3