Question

Tổng sau có chia hết cho 3 không ?

            \(A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}\)

Answer 1

Ta có 2A=2²+2³+...+2¹¹

=>2A-A=(2²+2³+...+2¹¹)-(2+2²+2³+...+2¹⁰)

=>A=2¹¹-2=2.(2¹⁰-1)

=2.1024-1=2.1023

Do 1023 chia hết cho 3=>A chia hết cho 3

Answer 2

A có 10 số hạng, ta chia A thành 5 nhóm, mỗi nhóm có 2 số hạng như sau:
A = 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰
= (2 + 2²) + (2³ + 2⁴) + (2⁵ + 2⁶) + (2⁷ + 28) + (2⁹ + 2¹⁰)
= 2(1 + 2) + 2³(1 + 2) + 2⁵(1 + 2) + 2⁷(1 + 2) + 2⁹(1 + 2)
= 2.3 + 2³.3 + 2⁵.3 + 2⁷.3 + 2⁹.3
= (2 + 2³ + 2⁵ + 2⁷ + 2⁹).3 \(⋮\) 3

Vậy A \(⋮\) 3

Answer 3

Ta có:

$A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}$

$=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\left(2^7+2^8\right)$

$+\left(2^9+2^{10}\right)$

$=2.\left(1+2\right)+2^3.\left(1+2\right)+2^5.\left(1+2\right)+2^7.\left(1+2\right)$

$+2^9.\left(1+2\right)$

= $2.3+2^3.3+2^5.3+2^7.3+2^9.3$

= $3.(2+2^3+2^5+2^7+2^9)$

Vậy A $⋮\; 3$