Question

CMR

(1+2+2²+2³+...+2¹¹) chia hét cho 9

Answer 1

Tui biết làm nè nhưng một lúc nữa nhé

Answer 2

Đặt \(A=1+2+2^2+2^3+...+2^{11}\)

\(A=\left(1+2+2^2+2^3+2^4+2^5\right)+2^6\left(1+2+2^2+2^3+2^4+2^5\right)\)

\(A=63+2^6.63\)

\(A=63\left(1+2^6\right)\)

Vì \(63⋮9\) nên \(63\left(1+2^6\right)⋮9\)

Vậy \(A⋮9\)

Answer 3

Đặt A= 1+2+2\(^{^{^{ }}2}\)+2\(^3\)+...+2\(^{11}\)

=> A=(1+2+2\(^{^{^{ }}2}\)+2\(^3\)+2\(^4\)+2\(^5\))+2\(^6\)(1+2+2\(^{^{^{ }}2}\)+2\(^3\)+2\(^4\)+2\(^5\))

=> A=63+2\(^6\).63

=> A=63.(1+2\(^6\))

Vì 63\(⋮\)9=>A chia hết cho 9