Question

Chứng minh rằng A=(1+2+2^2+2^3+......+2^99) chia hết cho 3

 

Answer 1

\(\left(1+2+2^2+2^3+...+2^{99}\right)\) )

= \(\left(1+2\right)+2^2\left(1+2\right)+2^3\left(1+2\right)+2^{98}\left(1+2\right)\)

=\(3+2^2.3+2^3.3+...+2^{98}.3\)\(⋮3\)(đccm)

Answer 2

A= 1+2+2²+...+2⁹⁹

A=( 1+2)+(2²+2³)+....+(2⁹⁸+2⁹⁹)

A=1(1+2)+ 2²(1+2)+....+2⁹⁸(1+2)

A= 1.3 + 2^2..3+.......+2⁹⁹.3

A= 3 ( 1+2²+....+2⁹⁹)

Vì 3:3 nên 3 3 ( 1+2²+....+2⁹⁹) chia hết cho 3

Vậy...

Answer 3

Ta có :\(A=1+2+2^2+2^3+....+2^{98}+2^{99}\)

              \(=\left(1+2\right)+\left(2^2+2^3\right)+\left(2^4+2^5\right)+....+\left(2^{98}+2^{99}\right)\)

              \(=\left(1+2\right)+2^2.\left(1+2\right)+2^4.\left(1+2\right)+....+2^{98}.\left(1+2\right)\)

              \(=3+2^2.3+2^4.3+...+2^{98}.3\)

              \(=3.\left(1+2^2+2^4+...+2^{98}\right)\) chia hết cho 3

\(\implies\) \(A\) chia hết cho 3

Answer 4

:v Đăng sau thì lại được k :vv

Answer 5

Bạn đang bảo mình à :<???