Ta thấy : \(n\inℤ^+\Rightarrow n=k+1\left(k\inℕ\right)\)
Khi đó : \(A=2^{3\left(k+1\right)+1}+2^{3\left(k+1\right)-1}+1\)
\(=2^{3k+4}+2^{3k+2}+1\)
\(=8^k.16+8^k.4+1\equiv1.2+1.4+1\equiv0\left(mod7\right)\)
Do vậy : \(A⋮7\) mà \(A>7\forall n\inℤ^+\)
\(\Rightarrow\)\(A=2^{3n+1}+2^{3n-1}+1\) là hợp số (đpcm)