Toán chứng minh đây (part 4)

Question

Cho A=2+2^2+...+2^2004

CMR: A chia hết cho 7.

Answer 1

A=2+2^2+...+2^2004

=> A=(2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^2002+2^2003+2^2004)

=> A= 2(1+2+2^2)+ 2^4(1+2+2^2)+...+  2^2002(1+2+2^2)

=> A= 2(1+2+4)+2^4(1+2+4)+...+2^2002(1+2+4)

=> A= 2.7+2^4.7+...+2^2002.7

=> A= (2+2^4+...+2^2002).7

Mà: 7 chia hết cho 7

=> (2+2^4+...+2^2002).7 chia hết cho 7

=> A chia hết cho 7

=> ĐPCM.

Comments

thank you very much !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

A= (2+2^2+2^3)+.......+(2^2002+2^2003+2^2004)

A=2(1+2+2^2)+.......+2^2002(1+2+2^2)

A=2*7+.............+2^2002*7

A=7(2+.........+2^2002)

Vậy A chia hết cho 7

XD

Answer 3

Vì  nên . Vậy 

Answer 4

A=2+2^2+2^3+2^4+2^5+2^6...+2^1999+2^2000+2^2001+2^2002+2^2003+2^2004

=2.(1+2+2^2)+2^4.(1+2+2^2)+...+2^1999.(1+2+2^2)+2^2002.(1+2+2^2)

=2.7+2^4.7+...+2^1999.7+2^2002.7

=7.(2+2^4+...+2^1999+2^2002)

=> A chia hết cho 7