Câu hỏi của Nguyễn Trọng Phúc

Question

cho A=2 + 2² + 2³+ ... + 2⁴⁰

^{CMR: A chia hết cho 5}

huỳnh hào dương · Accepted Answer

Ta có$A=2+2^2+2^3+...+2^{40}$

$=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{37}+2^{38}+2^{39}+2^{40}\right)$

$=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{37}\left(1+2+2^2+2^3\right)$

$=2.15+2^5.15+...+2^{37}.15=15.\left(2+2^5+...+2^{37}\right)⋮5$

Vậy $A⋮5$

Câu trả lời:

Ta có$A=2+2^2+2^3+...+2^{40}$

$=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{37}+2^{38}+2^{39}+2^{40}\right)$

$=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{37}\left(1+2+2^2+2^3\right)$

$=2.15+2^5.15+...+2^{37}.15=15.\left(2+2^5+...+2^{37}\right)⋮5$

Vậy $A⋮5$