a) Ta thấy: 2 + 2² + 2³ + 2⁴ chia hết cho 6

suy ra tổng trên chia hết cho 6

suy ra đpcm

2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰

=(2+2²+2³+2⁴)+...(2⁹⁷+2⁹⁸+2⁹⁹+2¹⁰⁰)

=2(1+2+2²+2³)+...+2⁹⁷(1+2+2²+2³)

=2.15+....+2⁹⁷.15

=15(2+...+2⁹⁷)

=> 2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰ chia hết cho 15 (1)

Ta có: 2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰ >2

^=> 2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰ chia hết cho 2 (2)

Từ (1) và (2) => 2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰ chia hết cho 30

=> đpcm

A= 2+2^2+2^3+...+2^2004. Chứng minh rằng : A chia hết cho 6

\(2+2^2+2^3+...+2^{2004}\)
\(=\left(2+2^2+2^3+2^4\right)\)\(+\left(2^5+2^6+2^7+2^8\right)\)\(+...+\left(2^{2001}+2^{2002}+2^{2003}+2^{2004}\right)\)
\(=2^0\left(2+2^2+2^3+2^4\right)\)\(+2^4\left(2+2^2+2^3+2^4\right)\)\(+...+2^{2000}\left(2+2^2+2^3+2^4\right)\)
\(=\left(2+2^2+2^3+2^4\right).\)\(\left(1+2^4+2^8+...+2^{2000}\right)\)
\(=30\left(1+2^4+2^8+...+2^{2000}\right)⋮30\)

ai biet giup

Bài 1

a) 3⁴ + 3⁵ + 3⁶ + 3⁷ = 3⁴(1 + 3 + 3² + 3³)\

b) a)A = 1 + 3 + 3² +......3⁹⁹ =(1 + 3 +  3² + 3³ ) + ...+(3⁹⁶ + 3⁹⁷ + 3⁹⁸ + 3⁹⁹)

                                          =   (1 + 3 +  3² + 3³ ) + .. +3⁹⁶(1 + 3 +  3² + 3³ )

                                          = 40 + ... + 3⁹⁶ . 40 

                                          = 40 (1 + 3 +...+ 3⁹⁶) chia hết cho 40

Bài 2 

a)

+)A chia hết cho 6

\(A=5+5^2+5^3+...+5^{2004}\)

\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{2003}+5^{2004}\right)\)

\(A=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{2002}\left(5+5^2\right)\)

\(A=30+5^2.30+...+5^{2002}.30\)

\(A=30\left(1+5^2+...+5^{2002}\right)\)chia hết cho 6

+)A chia hết cho 31

\(A=5+5^2+5^3+...+5^{2004}\)

\(A=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{2002}+5^{2003}+5^{2004}\right)\)

\(A=\left(5+5^2+5^3\right)+5^3\left(5+5^2+5^3\right)+...+5^{2001}\left(5+5^2+5^3\right)\)

\(A=155+5^3.155+...+5^{2001}.155\)

\(A=155\left(1+5^3+...+5^{2001}\right)\)chia hết cho 31

+) A chia hết cho 156

\(A=5+5^2+5^3+...+5^{2004}\)

\(A=\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+...+\left(5^{2001}+5^{2002}+5^{2003}+5^{2004}\right)\)

\(A=\left(5+5^2+5^3+5^4\right)+5^4\left(5+5^2+5^3+5^4\right)+...+5^{2000}\left(5+5^2+5^3+5^4\right)\)

\(A=780+5^4.780+...+5^{2000}.780\)

\(A=780\left(1+5^4+...+5^{2000}\right)\)chia hết cho 156

b)B=165+2^15 chia hết cho 33

ta có 165 chia hết cho 33

mà 2¹⁵ ko chia hết cho 33

vậy 165+2^15 không chia hết cho 33 hay B không chia hết cho 33.

chứng tỏ A= 1+\(3^1\)+\(3^2\)+....+\(3^{99}\)là B(4) và là B (40).

\(A=2^0+2^1+2^2+...+2^{2004}\)

\(A=\left(2^0+2^1+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8+2^9\right)+...+\left(2^{2000}+2^{2001}+2^{2002}+2^{2003}+2^{2004}\right)\)

\(A=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\cdot\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{2000}\cdot\left(2^0+2^1+2^2+2^3+2^4\right)\)

\(A=31+2^5\cdot31+...+2^{2000}\cdot31\)

\(A=31\cdot\left(1+2^5+...+2^{2000}\right)\)

\(\Rightarrow A⋮31\left(đpcm\right)\)

A =   (2004 + 2004² ) + ( 2004³ + 2004⁴)+  (2004⁵ + 2004⁶)  +............................+ (2004⁸ + 2004¹⁰)
A = 2004 ( 1 + 2004 ) + 2004³ ( 1 +2004 ) + .... + 2004⁸ ( 1+ 2004 )
A = 2004.2005 + 2004³.2005 +....+2004⁸.2005
A = 2005.(  2004 + 2004² + 2004³ + 2004⁴ +  2004⁵ + 2004⁶  +............................+ 2004⁸ + 2004¹⁰   )
Vậy A chia hết cho 2005

có sai đề ở chỗ 2004^8+2004^10 ko bn