Question

[Phạm Ngọc Thùy Dương]

cho S=\(1+2+2^2+2^3+2^4+2^5+2^6+2^7\)

chứng tỏ ràng S chia hết cho 3

Answer 1

[Nguyễn Vũ Minh Hiếu]

S = 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7

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

S =      3       + 2^2 . ( 1 + 2 ) + 2^4 . ( 1 + 2 ) + 2^6 . ( 1 + 2 )

S =      3       + 2^2 .      3      + 2^4 .       3      + 2^6 .    3

S =      3 . ( 2^2 + 2^4 + 2^6 )

Vi 3 chia het cho 3 nen 3 . ( 2^2 + 2^4 + 2^6 ) chia het cho 3

hay S chia het cho 3

Answer 2

[Ngọc_Hà]

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

\(\Rightarrow S=\)\(S=(1+2)+(2^2+2^3)+(2^4+2^5)+(2^6+2^7)\)

\(\Rightarrow S=\left(1+2\right)+2^2\left(1+2\right)+2^4\left(1+2\right)+2^6\left(1+2\right)\)

\(\Rightarrow S=3\cdot\left(1+2^2+2^4+2^6\right)⋮3\)

VẬY \(S⋮3\left(đpcm\right)\)