Question

Cmr : A chia hết cho 3 biest A = 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^9 + 2^10

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Answer 1

\(\text{​​}A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\left(2^7+2^8\right)+\left(2^9+2^{10}\right)\)

     \(=2.\left(1+2\right)+2^3.\left(1+2\right)+2^5.\left(1+2\right)+2^7.\left(1+2\right)+2^9.\left(1+2\right)\)

      \(=2.3+2^3.3+2^5.3+2^7.3+2^9.3\)

       \(=3\left(2+2^3+2^5+2^7+2^9\right)\)

Thấy : \(2+2^3+2^5+2^7+2^9\in N\)

\(\Rightarrow3\left(2+2^3+2^5+2^7+2^9\right)⋮3\)

Hay : \(A⋮3\)( đpcm )

Answer 2

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

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

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

A = 2.3 + 2³.3 + 2⁵. 3 + 2⁷.3 + 2⁹.3

A = 3.( 2 + 2³ + 2⁵ + 2⁷ + 2⁹ ) \(⋮\)3

Vậy A chia hết cho 3.