Question

Tổng sau có chia hết cho 5 ko

A = 1+2+2²+2³⁺2⁴+2⁵+2⁶+2⁷

^{Giair thích chi tiết giùm mk nha}

Answer 1

A = 1+2+2²+2³+2⁴+2⁵+2⁶+2⁷

A = (1+2+2²+2³) + (2⁴+2⁵+2⁶+2⁷)

A = 1(1+2+2²+2³) + 2⁴(1+2+2²+2³)

A = 1.15 + 2⁴.15

A = 15(1+2⁴)

Vì 15(1+2⁴) \(⋮\)5 nên A \(⋮\)5

Vậy A \(⋮\)5

Answer 2

Ta có:

A = 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷

A = (1 + 2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶ + 2⁷)

A = 1(1 + 2 + 2² + 2³) + 2⁴(1 + 2 + 2² + 2³)

A = 1.15 + 2⁴.15

A = (1 + 2⁴)15

Vì 15$⋮5\; nên$(1 + 2⁴)15$⋮5$

$=>\; A⋮5$

$Vậy\; A⋮5$

Answer 3

1+2+2²+2³+2⁴+2⁵+2⁶+2⁷

=(1+2+2²+2³)+(2⁴+2⁵+2⁶+2⁷)

=1. (1+2+2²+2³)+2⁴. (1+2+2²+2³)

=(1+2⁴) . (1+2+2²+2³)

=(1+2⁴) .15

=(1+2⁴). 5 .3 chia hết cho 5