\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^{n+2}+3^n-\left(2^{n+2}+2^n\right)\)
\(=3^n.3^2+3^n-\left(2^n.2^2+2^n\right)\)
\(=3^n\left(3^2+1\right)-2^n.\left(2^2+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10\)
\(=\left(3^n-2^{n-1}\right).10\) chia hết cho 10
Bảo nè,phải sửa lại đề n\(\in\)N* vì n=0 thì \(2^{0-1}=2^{-1}=\frac{1}{2}\) nên \(\left(3^n-2^{n-1}\right).10\) không chia hết cho 10
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.9-2^{n-1}.8+3^n-2^{n-1}.2\)
\(=3^n\left(9+1\right)-2^{n-1}\left(8+2\right)\)
\(=3^n.10-2^{n-1}.10\)
\(=10\left(3^n-2^{n-1}\right)\)\(⋮\)\(10\)
ta có: 3n+2 - 2n+2+3n-2n
= 3n.32-2n.22+3n-2n
= 3n.(32+1) - 2n.(22+1)
= 3n. 10 - 2n.5
= 3n.10 - 2n-1. 2.5
= 3n.10 - 2n-1.10
= 10.(3n-2n-1) chia hết cho 10
=> đ p c m
