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

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)

mn mn ơiii

helllppppppppp

\(2,\\ 3^{n-3}+2^{n-3}+3^{n+1}+2^{n+2}\\ =3^{n-3}\left(1+3^4\right)+2^{n-3}\left(1+2^5\right)\\ =3^{n-3}\cdot82+2^{n-3}\cdot33\)

Vì \(3^{n-3}\cdot82⋮2;⋮3\) nên \(3^{n-3}\cdot82⋮6\)

\(2^{n-3}\cdot33⋮2;⋮3\) nên \(2^{n-3}\cdot33⋮6\)

Do đó tổng trên chia hết cho 6 với mọi \(n\in N\)

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 $1$

\(B=1+3^1+3^2+...+3^{2016}\)

\(3B=3+3^2+3^3+3^4+...+3^{2017}\)

\(3B-B=3^{2017}-1\)

\(B=\dfrac{3^{2017}-1}{2}\)

\(B=1+3^1+3^2+...+3^{2016}\)

\(3\cdot B=3+3^2+3^3+...+3^{2016}+3^{2017}\)

\(3B-B=3+3^2+3^3+...+3^{2016}+3^{2017}-\left(1+3^1+3^2+...+3^{2016}\right)\)

\(2B=3^{2017}-1\)

\(\Rightarrow B=\dfrac{3^{2017}-1}{2}\)

\(B=1+3+3^2+...+3^{2016}\)

\(3\cdot B=3+3^2+3^3+...+3^{2017}\)

\(3B-B=3+3^2+3^3+...+3^{2017}-\left(1+3+3^2+...+3^{2016}\right)\)

\(2B=3^{2017}-1\)

\(\Rightarrow B=\dfrac{3^{2017}-1}{2}\)