Số số hạng của A:

90 - 1 + 1 = 90 (số)

Do 90 chia hết cho 3 nên có thể nhóm thành nhóm 3 số hạng

Ta có:

A = 2¹ + 2² + 2³ + ... + 2⁹⁰

= (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2⁸⁸ + 2⁸⁹ + 2⁹⁰)

= 2.(1 + 2 + 2²) + 2⁴.(1 + 2 + 2²) + ... + 2⁸⁸.(1 + 2 + 2²)

= 2.7 + 2⁴.7 + ... + 2⁸⁸.7

= 7.(2 + 2⁴ + ... + 2⁸⁸) ⋮ 7

Vậy A ⋮ 7

b) A = 2¹ + 2² + 2³ + ... + 2⁹⁰

⇒ 2A = 2² + 2³ + 2⁴ + ... + 2⁹¹

⇒ A = 2A - A = (2² + 2³ + 2⁴ + ... + 2⁹¹) - (2 + 2² + 2³ + ... + 2⁹⁰)

= 2⁹¹ - 2

Số số hạng:

(290-21):1+1=270( số hạng)

Tổng A:(290+21) x 270:2=41985

Ta có:41986:7 hết nén A chia hết cho 7.

Ta có : 

\(A=2+2^2+2^3+2^4...2^{2010}\)\(^0\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

\(=2.3+2^3.3+....+2^{2009}.3\)

\(=3\left(2+2^3+....+2^{2009}\right)⋮3\)

Ta có :

\(2+2^2+2^3+2^4+....+2^{2010}\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)

\(=2.7+2^4.7+....+2^{2008}.7\)

\(=7\left(2+2^4+....+2^{2008}\right)⋮7\)

Vậy \(2^1+2^2+2^3+2^4+...+2^{2010}⋮3\) và \(7\)

*Sửa lại đề*

A = 2¹+ 2²+ 2³+ 2⁴ + .. + 2¹⁰⁰

A = (2¹+2²) + (2³+ 2⁴) +...+ (2⁹⁹+ 2¹⁰⁰)

A = 2.(1+2) + 2³.(1+2) + .. + 2⁹⁹. (1+2)

A = 2.3 + 2³. 3 + .. + 2⁹⁹.3

A = 3 . (2¹ + 2³ + .... + 2⁹⁹)

Mà 3 chia hết cho 3 

=> A chia hết cho 3

A = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ … + 2⁹⁹

^A=( 2⁰ + 2¹ + 2² + 2³ + 2⁴⁾ +( 2⁵ … + 2⁹⁹⁾

A= 2⁰.(2⁰ + 2¹ + 2² + 2³ + 2⁴)+2⁵.( 2⁵ … + 2⁹⁹⁾

A= 1. 31+ 2⁵.31… + 2⁹⁵.31

^A= 31. (1+2⁵...+2⁹⁵)

KL: ...... 

\(A=2^0+2^1+2^2+2^3+2^4+...+2^{99}=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{95}\left(2^0+2^1+2^2+2^3+2^4\right)=31+31.2^5+...+31.2^{95}=31\left(1+2^5+...+2^{95}\right)⋮31\)

A = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ … + 2⁹⁹

A=( 2⁰ + 2¹ + 2² + 2³ + 2⁴⁾ +( 2⁵ … + 2⁹⁹⁾

A= 2⁰.(2⁰ + 2¹ + 2² + 2³ + 2⁴)+2⁵.( 2⁵ … + 2⁹⁹⁾

A= 1. 31+ 2⁵.31… + 2⁹⁵.31

A= 31. (1+2⁵...+2⁹⁵)

KL: ...... 

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=14+2^3\cdot14+...+2^{117}\cdot14\)

\(=14\cdot\left(1+2^3+...+2^{117}\right)⋮7\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=62+2^5\cdot62+...+2^{115}\cdot62\)

\(=62\cdot\left(1+2^5+...+2^{115}\right)⋮31\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5+2^6\right)+\left(2^7+2^8+2^9+2^{10}+2^{11}+2^{12}\right)+...+\left(2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=126+126\cdot2^6+...+126\cdot2^{114}\)

\(=126\cdot\left(1+2^6+...+2^{114}\right)⋮21\)

c) \(55-7.\left(x+3\right)=6\)

\(7.\left(x+3\right)=55-6\)

\(7.\left(x+3\right)=49\)

\(x+3=49:7\)

\(x+3=7\)

\(x=7-3\)

\(x=4\)

d) \(-14-x+\left(-15\right)=-10\)

\(-29-x=-10\)

\(x=-29+10\)

\(x=-19\)

-----------------------------

Số số hạng của A:

\(60-1+1=60\) (số)

Do \(60⋮6\) nên ta có thể nhóm các số hạng của A thành từng nhóm mà mỗi nhóm có 6 số hạng như sau:

\(A=\left(2+2^2+2^3+2^4+2^5+2^6\right)+\left(2^7+2^8+2^9+2^{10}+2^{11}+2^{12}\right)+...+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4+2^5\right)+2^7.\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}.\left(1+2+2^2+2^3+2^4+2^5\right)\)

\(=2.63+2^7.63+...+2^{55}.63\)

\(=63.\left(2+2^7+...+2^{55}\right)\)

\(=21.3.\left(2+2^7+...+2^{55}\right)⋮21\)

Vậy \(A⋮21\)

55-7(x+3)=6

7(x+3)=55-6=49

(x+3)=49:7=7

x=7-3=4

(-14)-x + (-15)=-10

(-14)-x=-10-15=-25

x           =-14-25=-39 

A chia hết 31 chứ

Câu 1: 

$A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+....+(2^{2019}+2^{2020})$

$=2(1+2)+2^3(1+2)+2^5(1+2)+....+2^{2019}(1+2)$

$=(1+2)(2+2^3+2^5+...+2^{2019})=3(2+2^3+2^5+...+2^{2019})\vdots 3$

-----------------

$A=2+(2^2+2^3+2^4)+(2^5+2^6+2^7)+....+(2^{2018}+2^{2019}+2^{2020})$

$=2+2^2(1+2+2^2)+2^5(1+2+2^2)+....+2^{2018}(1+2+2^2)$

$=2+(1+2+2^2)(2^2+2^5+....+2^{2018})$

$=2+7(2^2+2^5+...+2^{2018})$

$\Rightarrow A$ chia $7$ dư $2$.

Câu 2:

$B=(3+3^2)+(3^3+3^4)+....+(3^{2021}+3^{2022})$
$=3(1+3)+3^3(1+3)+...+3^{2021}(1+3)$

$=(1+3)(3+3^3+...+3^{2021})=4(3+3^3+....+3^{2021})\vdots 4$

-------------------

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^{2020}+3^{2021}+3^{2022})$

$=3(1+3+3^2)+3^4(1+3+3^2)+....+3^{2020}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+...+3^{2020})=13(3+3^4+...+3^{2020})\vdots 13$ (đpcm)