a) \(33^{n+1}-33^n=33^n.33-33^n\)
\(=33^n\left(33-1\right)=33^n.32\)
Vì \(32⋮32\forall n\) nên \(33^n.32⋮32\forall n\)
Vậy \(33^{n+1}-33^n⋮32\left(đpcm\right)\)
b) \(\left(4n+7\right)^2-49=\left(4n+7\right)^2-7^2\)
\(=\left(4n+7-7\right)\left(4n+7+7\right)=4n\left(4n+14\right)\)
\(=8n^2+64n=8\left(n^2+8n\right)\)
Vì \(8⋮8\forall n\) nên \(8\left(n^2+8n\right)⋮8\forall n\)
Vậy \(\left(4n+7\right)^2-49⋮8\forall n\left(đpcm\right)\)
Đúng 0
Bình luận (0)
