Question

Chung minh tong sau chia het cho7

A=2^1+2^2+2^3+....+2^59+2^60

Answer 1

A = (2¹ + 2² + 2³ + ... + 2⁵⁹ + 2⁶⁰)

A = 2⁰(2¹ + 2² + 2³) + 2³(2¹ + 2² + 2³) + ... + 2⁵⁷(2¹ + 2² + 2³)

A = (2¹ + 2² + 2³)  + (2⁰ + 2³ + ... + 2⁵⁷)

A = 14(2⁰ + 2³ + ... + 2⁵⁷) chia hết cho 7

=> A chia hết cho 7.

Answer 2

A=2(1+2+4)+2⁴(1+2+4)+2⁷(1+2+4)+...+2⁵⁸(1+2+4)
A=(1+2+4)(2+2⁴+2⁷+...+2⁵⁸)
A=7.(2+2⁴+2⁷+...+2⁵⁸) chia hết cho 7

Answer 3

A= (2^1 + 2^2 + 2^3) + (2^4 + 2^5 + 2^6) +...+ (2^58 + 2^59 + 2^60)

  = 2(1+2+2^2) + 2^4(1+2+2^2) +...+ 2^58(1+2+2^2)

  = 2.7 + 2^4.7 +...+ 2^58.7

  = 7(2 + 2^5 +...+ 2^58)

Vậy A chia hết cho 7