Question

H = 2+2^2 + 3^2  + ... + 2^60  chưng minh H chia hết cho 3 , 7 , 15 
Ai làm nhanh mình tick cho 

Answer 1

Có thể là bạn đã vt nhầm ở chỗ 3^2, đáng nhẽ ra là 2^3

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

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

\(H=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(H=2.3+2^3.3+...+2^{59}.3\)

\(H=3\left(2+2^3+5^{59}\right)\)

Có :\(H=3\left(2+2^3+5^{59}\right)⋮3\)

\(\Rightarrow H⋮3\)

Answer 2

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

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

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

\(H=2.7+2^4.7+...+2^{58}.7\)

\(H=7\left(2+2^4+...+2^{58}\right)\)

Có : \(H=7\left(2+2^4+...+2^{58}\right)⋮7\)

\(\Rightarrow H⋮7\)

Answer 3

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

\(H=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

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

\(H=2.15+...+2^{57}.15\)

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

Có : \(H=15\left(2+2^5+...+2^{57}\right)⋮15\)

\(\Rightarrow H⋮15\)