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

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

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

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

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

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

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

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

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

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 $1$

`#3107.101107`

\(S=1+3^1+3^2+3^3+...+3^{101}\)

\(3S=3+3^2+3^3+...+3^{102}\)

\(3S-S=\left(3+3^2+3^3+...+3^{102}\right)-\left(1+3+3^2+...+3^{101}\right)\)

\(2S=3+3^2+3^3+3^{102}-1-3-3^2-...-3^{101}\)

\(2S=3^{102}-1\)

\(S=\dfrac{3^{102}-1}{2}\)

Vậy, \(S=\dfrac{3^{102}-1}{2}.\)

3s=3+3^2+3^3+....+3^102

3s-s=2s

2s=3^102-1

s=3^102-1 trên2

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)

\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)

\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)