Bài 210 trang 32 SBT toán 6 tập 1

Đề bài

Tổng sau có chia hết cho \(3\) không\(?\)

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

Lời giải

Ta có:

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

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

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

\(= 2.3 + {2^3}.3 + {2^5}.3 + {2^7}.3 + {2^9}.3\)

\(= 3.(2 + {2^3} + {2^5} + {2^7} + {2^9})\)            

Vậy \(A\;\vdots\; 3\)


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