3n+2 - 2n+2 + 3n - 2n
= 3n+2 + 3n - 2n+2 + 2n
= 3n . (32 + 1) - 2n . (22 + 1)
= 3n . 10 - 2n . 5
= 3n . 10 - 2n-1 . 5 . 2
= 3n . 10 - 2n-1 . 10
= 10 . (3n - 2n-1)
=> 3n+2 - 2n+2 + 3n - 2n ⋮ 10
Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^2\cdot3^n+3^n-2^2\cdot2^n-2^n\)
\(=3^n\left(9+1\right)-2^n\left(2^2+1\right)\)
\(=10\cdot3^n-2^n\cdot5\)
\(=10\cdot3^n-2^{n-1}\cdot2\cdot5\)
\(=10\cdot3^n-10\cdot2^{n-1}\)
\(=10\cdot\left(3^n-2^{n-1}\right)⋮10\forall n\in N\)*(đpcm)
Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n.\left(3^2+1\right)-2^n.\left(2^2+1\right)\)
\(=3^n.\left(9+1\right)-2^n.\left(4+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n-2^{n-1}\right)⋮10\)
Vậy: \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
