Question

[Lê Nguyễn Ngọc Nhi]

Chứng tỏ rằng A chia hết cho 155, biết:

A= 2+2²+2³+2⁴+...+2⁹⁹+2^100.

^{Giúp mk nha}

Answer 1

[Hung nguyen]

Ta có: 155 = 5.31 ta chứng minh A chia hết cho 5 và 31

+ Chứng minh A chia hết cho 5

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(=2\left(1+2+4+8\right)+2^5\left(1+2+4+8\right)+...+2^{97}\left(1+2+4+8\right)\)

\(=15\left(2+2^5+...+2^{97}\right)=3.5.\left(2+2^5+...+2^{97}\right)\)

\(\Rightarrow A⋮5\left(1\right)\)

+ Chứng minh A chia hết cho 31

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(=2\left(1+2+4+8+16\right)+2^6\left(1+2+4+8+16\right)+...+2^{96}\left(1+2+4+8+16\right)\)

\(=31\left(2+2^6+...+2^{96}\right)\)

\(\Rightarrow A⋮31\left(2\right)\)

Từ (1) và (2) \(\Rightarrow A⋮\left(31.5\right)hayA⋮155\)