Nếu đúng là zậy thì mk biết làm.

A = 3 + 3² + 3³ + ...  + 3²⁰⁰⁴

A = (  3 + 3² + 3³ + 3⁴ ) + ... + ( 3²⁰⁰¹ + 3²⁰⁰² + 3²⁰⁰³ + 3²⁰⁰⁴ )

A = 3( 1 + 3 + 3² + 3³ ) + ... + 3²⁰⁰¹( 1 + 3 + 3² + 3⁹ )

A = 3.40 + ... + 3²⁰⁰¹.40

A = ( 3 + 3⁵ + ...  3²⁰⁰¹) . 40

=> A chia hết cho 40

A = 3 + 3² + 3³ +3⁴ + ... + 3²⁰⁰⁴ phải ko? 

sao lai la 3^9 vay bn

Bài 1:

a. $2^{29}< 5^{29}< 5^{39}$

$\Rightarrow A< B$

b.

$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$

$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$

$=(1+3)(3+3^3+3^5+...+3^{2009})$

$=4(3+3^3+3^5+...+3^{2009})\vdots 4$

Mặt khác:

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$

$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$

Bài 1:
c.

$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$

$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$

$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$

$\Rightarrow A=\frac{3^{101}+1}{4}$

Bài 2:

a. $7\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 7; -7\right\}$

$\Rightarrow n\in \left\{0; -2; 6; -8\right\}$

b.

$2n+5\vdots n+1$
$\Rightarrow 2(n+1)+3\vdots n+1$

$\Rightarrow 3\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow n\in \left\{0; -2; 2; -4\right\}$

a: \(A=2019\cdot2021=2020^2-1\)

\(B=2020^2\)

Do đó: A<B

Bài 1:

\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)

Bài 2:

\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)

Bài 1 :

\(2^{49}=\left(2^7\right)^7=128^7\)

\(5^{21}=\left(5^3\right)^7=125^7\)

mà \(125^7< 128^7\)

\(\Rightarrow2^{49}>5^{21}\)

Bài 2 :

a) \(S=1+3+3^2+3^3+...3^{99}\)

\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)

\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

b) \(S=1+4+4^2+4^3+...4^{62}\)

\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)

\(\Rightarrow S=21+4^3.21+...4^{60}.21\)

\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)

\(\Rightarrow dpcm\)

1. So sánh bằng cách nhanh nhất:

a. \(\frac{28}{31}>\frac{28}{33}\)

b.\(\frac{28}{81}>\frac{11}{34}\)

c. \(\frac{13}{11}>\frac{31}{32}\)

d. \(\frac{1994}{1995}>\frac{36}{37}\)

Giúp mình với!!!

giải cả cách làm b ơi

A=11+22+33+44

=33+77

=110

B=14+23+32+41

=37+73

=110

VÌ 110=110 NÊN A =B

A = 11 + 22 + 33 + 44

   = 33 + 33 + 44

   = 66 + 44

   = 110

B = 14 + 23 + 32 + 41

   = 37 + 73

   = 110

Vậy A = B ( = 110 )

\(A=11+22+33+44=10+1+20+2+30+3+40+4\)

\(A=10+4+20+3+30+2+40+1=14+23+32+41=B\)

a)  18 2 < 10 3

b)  3 2 + 4 2 < ( 3 + 4 ) 2

c)  100 2 + 30 2 < ( 100 + 30 ) 2

d)  a 2 + b 2 > ( a - b ) 2 với a ∈   N * ;   b ∈   N * .