Question

Chứng minh rằng với mỗi số nguyên dương n thì:

\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)

Answer 1

Ta có : \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)

\(=3^{n+1}\left(3^2+1\right)+2^{n+2}\left(2+1\right)\)

\(=3^{n+1}.10+2^{n+1}.3\)

\(=3^n.5.6+2^{n+1}.6⋮6\)

Answer 2

Ta có: \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)

\(=3^{n+2}.\left(3^2+1\right)+2^{n+2}.\left(2+1\right)\)

\(=3^{n+1}.10+2^{n+2}.3\)

\(=3^n.3.10+2^{n+1}.2.3\)

\(\Rightarrow3^n.5.6+2^{n+1}.6⋮6\)

\(\Rightarrow3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)