Question

[2TQEFSCF32]

giúp mình b2 với ạ, b1 mình lm r, cảm ơnnn!!

Answer 1

[HT.Phong (9A5)]

Bài 2: 

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

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

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

\(A=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)

\(A=3\cdot\left(2+2^3+...+2^{59}\right)\) ⋮ 3

Vậy: A ⋮ 3

_____________

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

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

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

\(A=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)

\(A=7\cdot\left(2+2^4+....+2^{58}\right)\) ⋮ 7

Vậy: A ⋮ 7

___________________

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

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

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

\(A=2\cdot5+2^2\cdot5+...+2^{58}\cdot5\)

\(A=5\cdot\left(2+2^2+...+2^{58}\right)\) ⋮ 5

Vậy: A ⋮ 5 

Answer 2

[2TQEFSCF32]

cảm ơnn