Bài 1 : chứng mình rằng : 2^2 + 2^3 + 2^4 +…+ 2^121 chia hết cho 31

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1. `2^2+2^3+2^4+…+2^121`

   `=(2^2+2^3+2^4+2^5+2^6)+(2^7+2^8+2^9+2^10+2^11)+..+(2^117+2^118+2^119+2^120+2^121)`

   `=2^2(1+2+4+8+16)+2^7(1+2+4+8+16)+…+2^117(1+2+4+8+16)`

   `=(2^2+2^7+2^12+…+2^117)(1+2+4+8+16)`

   `=(2^2+2^7+2^12).31`$\vdots$`31`

   `⇒(2^2+2^3+2^4+…+2^121)`$\vdots$`31`

    

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2. Ta có : `2^2 + 2^3 + 2^4 + … + 2^121`

   `= ( 2^2 + 2^3 + 2^4 + 2^5 + 2^6 ) + … + ( 2^117 + 2^118 + 2^119 + 2^120 + 2^121 )`

   `= 2^2 . ( 1 + 2 + 2^2 + 2^3 + 2^4 ) + … + 2^117 . ( 1 + 2 + 2^2 + 2^3 + 2^4 )`  

   `= ( 1 + 2 + 2^2 + 2^3 + 2^4 ) . ( 2^2 + …. + 2^116 )`

   `= 31 . ( 2^2 + … + 2^116 )` chia hết cho 31 ( Điều phải chứng minh )

    

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