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\(A=2+2^2+2^3+...+2^{10}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)
\(=3\left(2+2^3+...+2^9\right)⋮3\)
\(A=2+2^2+2^3+...+2^{10}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)\)
\(=\left(2+2^6\right).31⋮31\)
\(3+3^2+3^3+...+3^{2012}\)
\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=3\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)
\(=40\left(3+...+3^{2009}\right)⋮40\)
\(a=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)=\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)=\)
\(=3\left(2+2^3+2^5+2^7+...+2^{99}\right)⋮3\)
A=9^23 + 5 x 3^43
A=(3^2)^23 + 5 x 3 ^43
A=3^46+5x3^43
A=3^43(3^3+5)
A=3^43(27 + 5)
A=3^43x32
vì 32 chia hết cho 32
vậy A chia hết cho 32