Question

Bài 14 

A=2=2²+...+2⁶⁰

chứng tỏ a⋮3,a⋮7,a⋮15,a⋮21

Answer 1

Sửa đề: \(A=2+2^2+...+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{57}\right)\)

=>A chia hết cho 15 và A chia hết cho 3

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

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{58}\right)⋮7\)

Ta có: \(A⋮3;A⋮7\)

ƯCLN(3;7)=1

Do đó: \(A⋮3\cdot7=21\)