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\(vt=1+2015+2015^2+2015^3+2015^4+2015^5+2015^6+2015^7\)
\(=\left(1+2015\right)+\left(2015^2+2015^3\right)+\left(2015^4+2015^5\right)+\left(2015^6+2015^7\right)\)
\(=1\left(1+2015\right)+2015^2\left(1+2015\right)+2015^4\left(1+2015\right)+2015^6\left(1+2015\right)\)
\(=\left(2015+1\right)\left(1+2015^2+2015^4+2015^6\right)\)
\(=2016\left(1+2015^2+2015^4+2015^6\right)\)
\(=2016\left[\left(1+2015^2\right)+\left(2015^4+2015^6\right)\right]\)
\(=2016\left[1\left(1+2015^2\right)+2015^{2014}\left(1+2015^2\right)\right]=vp\left(đpcm\right)\)
\(=2016\left(1+2015^{2014}\right)\left(1+2015^{2012}\right)\)
Ta có \(2015^{2015}-2015^{2014}=2015^{2014}.2015-2015^{2014}=2015^{2014}.\left(2015-1\right)=2015^{2014}.2014\) chia hết cho 2014 (đpcm).
Cảm ơn bạn ạ