Với \(n=0\Rightarrow0-0+0-0+0-0=0⋮24\left(đúng\right)\)
Với \(n=1\Rightarrow1-3+6-7+5-2=0⋮24\left(đúng\right)\)
G/s \(n=k\Rightarrow\left(k^6-3k^5+6k^4-7k^3+5k^2-2k\right)⋮24\)
\(\Rightarrow k\left(k^5-3k^4+6k^3-7k^2+5k-2\right)⋮24\\ \Rightarrow k\left(k+1\right)\left(k^2+k+1\right)\left(k^2-k+2\right)⋮24\)
Với \(n=k+1\), ta cần cm \(\left[\left(k+1\right)^6-3\left(k+1\right)^5+6\left(k+1\right)^4-7\left(k+1\right)^3+5\left(k+1\right)^2-2\left(k+1\right)\right]⋮24\)
Ta có \(\left(k+1\right)^6-3\left(k+1\right)^5+6\left(k+1\right)^4-7\left(k+1\right)^3+5\left(k+1\right)^2-2\left(k+1\right)\)
\(=\left(k+1\right)\left[\left(k+1\right)^5-3\left(k+1\right)^4+6\left(k+1\right)^3-7\left(k+1\right)+5\left(k+1\right)-2\right]\\ =\left(k+1\right)\left(k+1-1\right)\left[\left(k+1\right)^2-\left(k+1\right)+1\right]\left[\left(k+1\right)^2-\left(k+1\right)+2\right]\\ =k\left(k+1\right)\left(k^2+k+1\right)\left(k^2+k+2\right)\)
Mà theo GT quy nạp ta có \(k\left(k+1\right)\left(k^2+k+1\right)\left(k^2+k+2\right)⋮24\)
Vậy ta được đpcm