2n+2+2n+1+2n=2n.(22+2+1)=2n.7
=> 2n+2+ 2n+1 + 2n chia hết cho 7 vs mọi n \(\in\)N
\(2^{n+2}+2^{n+1}+2^n=2^n.\left(2^2+2+1\right)=2^n.7\) chia hết cho 7
2n+2+2n+1+2n
= 2n^2+2n^1
Chia hết cho 7 về mọi £N
Ta có:
\(2^{n+2}+2^{n+1}+2^n=2^n\left(2^2+2+1\right)=2^n.7⋮7\)
Vậy...
\(2^{n+2}+2^{n+1}+2^n\)
\(=2^n\left(2^2+2+1\right)\)
\(=2^n.7⋮7\forall n\in N\)
