a: \(G=8^8+2^{20}\)

\(=2^{24}+2^{20}\)

\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)

b: Sửa đề: \(H=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{59}\right)⋮3\)

\(H=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{58}\right)⋮7\)

\(H=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{57}\right)⋮15\)

c: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)

\(E=1+3+3^2+3^3+...+3^{1991}\)

\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)

\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)

\(A=1+3+3^2+3^3+...+3^{2018}+3^{2019}\)

\(=\left(1+3\right)+3^2\left(1+3\right)+...+3^{2018}\left(1+3\right)\)

\(=\left(1+3\right)\left(1+3^2+...+3^{2018}\right)\)

\(=4\left(1+3^2+...+3^{2018}\right)\) ⋮4

⇒A⋮4

\(A=\left(1+3\right)+3^2\left(1+3\right)+...+3^{2018}\left(1+3\right)\)

\(=4\left(1+3^2+...+3^{2018}\right)⋮4\)

A=3²⁰¹⁹+1+3+3²+3³+...+3²⁰¹⁸

⇒A=1+3+3²+...+3²⁰¹⁸+3²⁰¹⁹ 

⇒3A=3×(1+3+3^2+3^3+....+3^2019)

3A=3+3^2+3^3+....+3^2020

3A-A=(3+3^2+3^3+....+3^2020) -(1+3+3^2+....+3^2019)

2A= 3^2020-1

⇒ A =( 3^2020-1):2

A=3²⁰¹⁹+1+3+3²+3³+...+3²⁰¹⁸

⇒A=1+3+3²+...+3²⁰¹⁸+3²⁰¹⁹ 

⇒3A=3×(1+3+3^2+3^3+....+3^2019)

⇒3A=3+3^2+3^3+....+3^2020

⇒3A-A=(3+3^2+3^3+....+3^2020) -(1+3+3^2+....+3^2019)

⇒2A= 3^2020-1

⇒ A =( 3^2020-1):2

\(A=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}\right)+\left(\frac{1}{46}+...+\frac{1}{60}\right)>\frac{1}{45}.15+\frac{1}{60}.15=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

=>đpcm

l-i-k-e cho mình nha

vì sao lại thế

\(16^{10}+32:3\)

\(16^{10}+32^7=\left(2^4\right)^{10}+\left(2^5\right)^7=2^{40}+2^{35}=2^{35}.2^5+3^{35}=2^{35}.\left(2^5+1\right)=2^{35}.33\)

chia hết cho 33

tick nhé

bn Hoàng Phúc làm đúng r đó

A=1+3+3^2+3^3+...+3^98+3^99+3^100

A=(1+3+ 3^2)+(3^3+3^4+3^5)+...+(3^98+3^99+3^100)

A=(1+3+3^2)+3^3x(1+3+3^2)+...+3^98x(1+3+3^2)

A=13x3^3x13+...+3^98x13

=> 13x(1+3+3^3+...+3^98)chia hết cho 13

Vậy A chia hết cho 13

câu b đâu bạn ?

cho A = 1 + 3 + 3² + 3³ + ... +  3¹¹

^b) chứng minh A chia hết cho 40

B=(16^10+32^7)

=(2^4)^10+(2^5)^7

=2^40+2^35

=2^35(2^5+1)

=2^35(32+1)

=2^35.33

=> B chia hết cho 33

=> 16^10+32^7 chia hết cho 33