Cho P=1+2+2^2+2^3+2^4+2^5+2^6+2^7 Chứng tỏ P chia hết cho 3

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1. $P=1^{}+$ $2^{2}+$ $2^{3}+$ $2^{4}+$ $2^{5}+$ $2^{6}+$ $2^{7}$ 

   $P=(1^{}+2)$ $(2^{2}+$ $2^{3})(+$ $2^{4}+$ $2^{5})+($ $2^{6}+$ $2^{7})$ 

   $\text{P = 1 .( 1 + 2 ) + }$ $2^{2}.(1+2)+$ $2^{4}.(1+2).$ $2^{6}.(1+2)$

   $P=1.3+2^{2}.3+$ $2^{4}.3+$ $2^{6}.3$

   $\text{P = 3 . ( 1 + }$ $2^{2}+$ $2^{4}+$ $2^{6})$\

   $\text{vậy P chia hết cho 3.}$

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2. P = 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7

   P = ( 1 + 2) + ( 2^2 + 2^3) + ( 2^4 + 2^5) + ( 2^6 + 2^7)

   P = 1 ( 1 + 2) + 2^2 ( 1 + 2) + 2^3 ( 1 + 2) + 2^4 ( 1 + 2) + 2^5 ( 1 + 2) + 2^6 ( 1 + 2)

   P = 1 . 3 + 2^2 . 3 + 2^3 . 3 + 2^4 . 3 + 2^5 . 3 + 2^6 . 3

   P = 3 ( 1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6)

   Mà 3 chia hết cho 3

   => 3 ( 1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6) chia hết cho 3

    

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