b) Ta có

     A = 3 + 3² + ... + 3²⁰⁰⁴.

=> A = 3 ( 1+ 3 + 3² ) + 3⁴  ( 1+ 3 + 3² ) + ... + 3²⁰⁰¹ ( 1+ 3 + 3² )

=> A = 3 . 13 + 3⁴ . 13 + ... + 3²⁰⁰¹ . 13

=> A = 13 ( 3 + 3⁴ + ... + 3²⁰⁰¹)  chia hết cho 13.

   Lại có :

     A = 3 + 3² + ... + 3^2004.

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

=> A = 3 ( 1+ 9) + 3² ( 1+ 9) + ... + 3²⁰⁰³ ( 1+ 9)

=> A = 10 ( 3 + 3² + ... + 3 ²⁰⁰³) chia hết cho 10.

 Vậy A vừa chia hết cho 13 vừa chia hết cho 10 mà ( 13;10) = 1

=> A chia hết cho 130.

A=3+3²+3³+......+3²⁰⁰⁴

3A=3²+3³+......+3²⁰⁰⁵

3A-A= ( 3²+3³+......+3²⁰⁰⁵ ) - ( 3+3²+3³+......+3²⁰⁰⁴ )

2A=3²⁰⁰⁵-3

A=\(\frac{3^{2005}-3}{2}\)

Ta có:

Ư(13)={1;13}

1)

a) $A=3+3^2+3^3+3^4+3^5+3^6+....+3^{28}+3^{29}+3^{30}$

$\iff A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+....+\left(3^{28}+3^{29}+3^{30}\right)$

$\iff A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+....+3^{28}\left(1+3+3^2\right)$

$\iff A=3.13+3^4.13+....+3^{28}.13$

$\iff A=13\left(3+3^4+....+3^{28}\right)⋮13\left(dpcm\right)$

b) $A=3+3^2+3^3+3^4+3^5+3^6+....+3^{25}+3^{26}+3^{27}+3^{28}+3^{29}+3^{30}$

$\iff A=\left(3+3^2+3^3+3^4+3^5+3^6\right)+....+\left(3^{25}+3^{26}+3^{27}+3^{28}+3^{29}+3^{30}\right)$

$\iff A=3\left(1+3+3^2+3^3+3^4+3^5\right)+....+3^{25}\left(1+3+3^2+3^3+3^4+3^5\right)$

$\iff A=3.364+....+3^{25}.364$

$\iff A=364\left(3+3^5+3^{10}+....+3^{25}\right)$

$\iff A=52.7\left(3+3^5+3^{10}+....+3^{25}\right)⋮52\left(dpcm\right)$

2) $A=3+3^2+3^3+....+3^{30}$

$\iff 3A=3\left(3+3^2+3^3+....+3^{30}\right)$

$\iff 3A=3^2+3^3+3^4+....+3^{30}+3^{31}$

$\iff 3A-A=\left(3^2+3^3+3^4+....+3^{30}+3^{31}\right)-\left(3+3^2+3^3+....+3^{30}\right)$

$\iff 2A=3^{31}-3$

$\iff A=\frac{3^{31}-3}{2}$

Vậy A không phải là số chính phương
Học tốt nha

a) $A=3+3^2+3^3+3^4+3^5+3^6+....+3^{28}+3^{29}+3^{30}$

$\iff A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+....+\left(3^{28}+3^{29}+3^{30}\right)$

$\iff A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+....+3^{28}\left(1+3+3^2\right)$

$\iff A=3.13+3^4.13+....+3^{28}.13$

$\iff A=13\left(3+3^4+....+3^{28}\right)⋮13\left(dpcm\right)$

b) $A=3+3^2+3^3+3^4+3^5+3^6+....+3^{25}+3^{26}+3^{27}+3^{28}+3^{29}+3^{30}$

$\iff A=\left(3+3^2+3^3+3^4+3^5+3^6\right)+....+\left(3^{25}+3^{26}+3^{27}+3^{28}+3^{29}+3^{30}\right)$

$\iff A=3\left(1+3+3^2+3^3+3^4+3^5\right)+....+3^{25}\left(1+3+3^2+3^3+3^4+3^5\right)$

$\iff A=3.364+....+3^{25}.364$

$\iff A=364\left(3+3^5+3^{10}+....+3^{25}\right)$

$\iff A=52.7\left(3+3^5+3^{10}+....+3^{25}\right)⋮52\left(dpcm\right)$

câu a, b trên mạng có nha 

c) do 3 +3^2+3^3+..+3^2004 chia hết cho 3

mà 3 ko chia hết cho 3^2 , 3^2 chia hét cho 3^2 ,.., 3^2004 chia hết cho 3^2 => a ko chia hết cho 3^2

=> a ko là scp ( do scp chie hết cho 3 , ko chia hết cho 3^2 , 3 nguyên tố)

Việt Hoàng sai

a)A=3^2005-1