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) \(S=1+4+4^2+4^3+...4^{62}\)

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

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

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

\(\Rightarrow dpcm\)

S = (1 - 3 + 3² - 3³) + 3⁴ . (1 - 3 + 3² - 3³) + .... + 3⁹⁶ . (1 - 3 + 3² - 3³)

S = (-20) + 3⁴ . (-20) +.... + 3⁹⁶ . (-20)

S = (-20) . (1 + 3⁴ +...+ 3⁹⁶) 

\(\Rightarrow\)S \(⋮\) 20 

(Ko bt có đúng ko)

*KO CHÉP MẠNG*

  a,

S  =     1 -  3 + 3² - 3³+...+3⁹⁸ - 3⁹⁹

S  =   3⁰ - 3¹ + 3² - 3³+...+ 3⁹⁸ - 3⁹⁹

xét dãy số: 0; 1; 2; 3;...;99 

Dãy số trên là dãy số cách đều với khoảng cách là: 1 - 0 = 1

Dãy số trên có số số hạng là: (99 - 0): 1 + 1 = 100 (số)

100 : 4 = 25

Vậy ta nhóm 4 số hạng liên tiếp của tổng S thành 1 nhóm thì: 

S = ( 1 - 3 + 3² - 3³) +....+( 3⁹⁶ - 3⁹⁷ + 3⁹⁸ - 3⁹⁹)

S = - 20+...+ 3⁹⁶.(1 - 3 + 3² - 3³)

S = - 20 +...+ 3⁹⁶.(-20)

S = -20.( 3⁰ + ...+ 3⁹⁶) (đpcm)

b,

  S           = 1 - 3 + 3² - 3³+...+ 3⁹⁸ - 3⁹⁹

3S          =      3  - 3² + 3³-...-3⁹⁸  + 3⁹⁹ - 3¹⁰⁰

3S + S   =    - 3¹⁰⁰ + 1

4S        = - 3¹⁰⁰ + 1 

 S        = ( -3¹⁰⁰ + 1): 4

S        = - ( 3¹⁰⁰ - 1) : 4

Vì S là số nguyên nên 3¹⁰⁰ - 1 ⋮ 4 ⇒ 3¹⁰⁰ : 4 dư 1 (đpcm)

Bài 1 :

a) \(a.b+b.19=713\) \(\left(a;b\inℕ^∗\right)\)

\(\Rightarrow b.\left(a+19\right)=713\)

\(\Rightarrow\left(a+19\right);b\in\left\{1;23;31;713\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(-18;713\right);\left(4;31\right);\left(12;23\right);\left(694;1\right)\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(4;31\right);\left(12;23\right);\left(694;1\right)\right\}\left(a;b\inℕ^∗\right)\)

b) \(a.b-10.b=650\)

\(\Rightarrow b.\left(a-10\right)=650\)

\(\Rightarrow\left(a-10\right);b\in\left\{1;5;10;13;25;26;50;65;130;325;650\right\}\)

Bạn lập bảng sẽ tìm ra (a;b)...

Bài 2 :

a) \(3^4+3^5+3^6+3^7=3^4\left(1+3+3^2+3^3\right)=3^4.40\)

b) \(B=1+3+3^2+3^3+...+3^{99}\)

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

\(\Rightarrow B=40+3^4.40...+3^{96}.40\)

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

\(\Rightarrow dpcm\)

Các bạn giúp mình nhé

\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)

\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)

\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)

\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)

\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)

\(S=4\left(3^2+3^4+3^6+3^8\right)\)

\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)

\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)

\(S=4x\left(1+3^2+...+3^8\right)\)

Vì 4 chia hết cho 4 nên S chia hết cho 4

\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)

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

\(=4.\left(3+3^3+...+3^{2009}\right)\)

⇒ \(B\) ⋮ 4

b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)