Question

Tong sau co chia het cho3 ko

A= 2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10

Answer 1

Có chia hết cho 3

Cứ ghép 2 số lại với nhau, sau đó nhân phân phói ra là xong : 

VD : \(2+2^2=2.1+2.2=2.\left(1+2\right)=2.3\)

Answer 2

A = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰

A = ( 2 + 2² ) + ( 2³ + 2⁴ ) + ... + ( 2⁹ + 2¹⁰ )

A = 2 . ( 1 + 2 ) + 2³ . ( 1 + 2 ) + ... + 2⁹ . ( 1 + 2 )

A = 2 . 3 + 2³ . 3 + ... + 2⁹ . 3

= ( 2 + 2³ + ... + 2⁹ ) . 3 \(⋮\)3

Answer 3

Ta có:A= 2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10=

= (2+2^2)+(2^3+2^4)+(2^5+2^6)+(2^7+2^8)+(2^9+2^10)=

= 6+2^2.(2+2^3)+2^4.(2+2^2)+2^6.(2+2^2)+2^8.(2+2^2)=

=6+ 2^2.6+2^4.6+2^6.6+2^8.6=6.(1+2^2+2^4+2^8)

Vì 6\(⋮\)3 nên 6.(1+2^2+2^4+2^8)\(⋮\)3 hay A\(⋮\)3

Answer 4

A = 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰

A = ( 2 + 2² ) + ( 2³ + 2⁴ ) + ( 2⁵ + 2⁶ ) + ( 2⁷ + 2⁸ ) + ( 2⁹ + 2¹⁰ )

A =  2 x ( 1 + 2 ) + 2³ x ( 1 + 2 ) + 2⁵ x ( 1 + 2 ) + 2⁷ x ( 1 + 2 ) + 2⁹ x ( 1 + 2 )

A =  2 x   3          + 2³ x   3          + 2⁵ x     3       +  2⁷ x   3         + 2⁹ x 3

Có:   2 x 3 \(⋮\)3

       2³ x 3 \(⋮\)3

       2⁵ x 3 \(⋮\)3

       2⁷ x 3 \(⋮\)3

       2⁹ x 3 \(⋮\)3

\(\Rightarrow\)A  \(⋮\)3