Question

Cho A = \(2+2^2+2^3+.....+2^{60}\). Chứng minh rằng A chia hết cho 3 ; 7 ; 15

Answer 1

A = 2 + 2² + 2³ +...+ 2⁶⁰

A = (2+2²) + (2³ + 2⁴) + ...+ (2⁵⁹ + 2⁶⁰)

A = 2.(1+2) + 2³.(1+2) + ...+ 2⁵⁹.(1+2)

A = 2.3 + 2³.3 + ....+ 2⁵⁹.3

A = 3.(2+2³ +...+2⁵⁹) chia hết cho 3

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các bài còn lại bn dựa zô mak lm\

Answer 2

\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(A=6+2^2\left(2+2^2\right)+...+2^{58}\left(2+2^2\right)\)

\(A=6\cdot1+2^2\cdot6+...+2^{58}\cdot6\)

\(A=6\cdot\left(1+2^2+...+2^{58}\right)⋮3\)

CMTT

Answer 3

very i dì