Question

Chứng minh rằng A = 2 + 2^2 + 2^3 +........+2 ^ 120  chia hết cho 7

Answer 1

A=2+2^2+2^3+...+2^120

A=(2+2^2+2^3)+(2^4+2^5+2^6)...+(2^118+2^119+2^120)

A=2.(1+2+2^2)+2^4(1+2+2^2)+2^118(1+2+2^2)

A=2.7+2^4.7+...+2^118.7

Ta có A=2.7+2^4.7+...+2^118.7 chia hết cho 7

=>A=2+2^2+2^3+...+2^120 chia hết cho 7

Answer 2

A=2+2^2+...+2^120

=(2+2^2+2^3+2^4+2^5+2^6)+(2^7+2^8+2^9+2^10+2^11+2^12)+.....+(2^120+2^119+2^118+2^117+2^116+2^115)

=2(1+2+2^2+2^3+2^4+2^5)+2^7(1+2+2^2+2^3+2^4+2^5)+.....+2^115(1+2+2^2+2^3+2^4+2^5)

=2*63+2^7*63+...+2^115*63

=63(2+2^7+...+2^115) Vì 63 chia hết cho 7=>63(2+2^7+..+2^115) chia hết cho 7

=>A chia hết cho 7

Answer 3

chung minh rang A=7^0+7^1+7^2+7^3+................+7^30+7^31 chia het cho 25