Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\) với mọi số nguyên dương n
\(=3^{n+2}+3^n-2^{n+2}-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=\left(3^n\times3^2+3^n\times1\right)-\left(2^n\times2^2+2^n\times1\right)\)
\(=\left\{3^n\times\left(3^2+1\right)\right\}-\left\{2^n\times\left(2^2+1\right)\right\}\)
\(=\left\{3^n\times\left(9+1\right)\right\}-\left\{2^n\times\left(4+1\right)\right\}\)
\(=3^n\times10-2^n\times5\)
\(=3^n\times10-2^{n-1+1}\times5\)
\(=3^n\times10-2^{n-1}\times2^1\times5\)
\(=3^n\times10-2^{n-1}\times2\times5\)
\(=3^n\times10-2^{n-1}\times10\)
\(=\left(3^n-2^{n-1}\right)\times10\)
Vì \(10⋮10\)
Nên \(\left(3^n-2^{n-1}\right)\times10⋮10\)
\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Vậy với mọi số nguyên dương n thì \(3^{n+2}-2^{n+2}+3^n-2^n\) chia hết cho 10 (đpcm)
