Cho A = 2^30 + 2^29 + 2^28
Chứng tỏ A chia hết cho 7.
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A=2+22+......+230
2A=22+23+............+ 231
A=(22+23+............+ 231)-(2+22+......+230)
A=22+23+............+ 231-2-22-...-230
A= 231-2
A=2.2.229-2
A=4.229-2
A\(⋮\)4 ( vì 4 chia hết cho 4 )
theo mk A ko chia hết cho 4
vì 22\(⋮\)4; 23\(⋮\)4 ;...
mà 2 ko chia hết cho 4 nên A ko chia hết cho 4
ta có
\(a+a^2+a^3+...+a^{30}\)
\(=a\left(1+a\right)+a^3\left(1+a\right)+a^5\left(1+a\right)+...+a^{29}\left(1+a\right)\)
\(=\left(a+a^3+a^5+...+a^{29}\right)\left(1+a\right)\)chia hết cho 1+a hay a=a^2+a^3+...+a^30 chia hết a+1 với a là số tự nhiên