Question

[Đỗ Nam Trâm]

S=\(2+2^2+2^3+...+2^{\text{2010}}\) chia hết cho 7,31

Answer 1

[Nguyễn Việt Lâm]

\(S=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{2008}.7\)

\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)

\(S=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{2006}\left(1+2+2^2+2^3+2^4\right)\)

\(=2.31+2^6.31+...+2^{2006}.31\)

\(=31\left(2+2^6+...+2^{2006}\right)⋮31\)

Answer 2

[Nguyễn Lê Phước Thịnh]

Ta có: \(S=2+2^2+2^3+...+2^{2010}\)

\(=\left(2+2^2+2^3+2^4+2^5+2^6\right)+...+\left(2^{2005}+2^{2006}+2^{2007}+2^{2008}+2^{2009}+2^{2010}\right)\)

\(=126\cdot\left(1+...+2^{2005}\right)⋮7\)

Ta có: \(S=2+2^2+2^3+...+2^{2010}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{2006}+2^{2007}+2^{2008}+2^{2009}+2^{2010}\right)\)

\(=31\cdot2\cdot\left(1+...+2^{2006}\right)⋮31\)