a: \(55^{n+1}-55^n\)
\(=55^n\cdot55-55^n\)
\(=55^n\left(55-1\right)=55^n\cdot54⋮54\)
b: \(n^2\left(n+1\right)+2n\left(n+1\right)\)
\(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
Vì n;n+1;n+2 là ba số nguyên liên tiếp
nên \(n\left(n+1\right)\left(n+2\right)⋮3!=6\)
=>\(n^2\left(n+1\right)+2n\left(n+1\right)⋮6\)
c: \(\left(2n-1\right)^3-\left(2n-1\right)\)
\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1\right]\)
\(=\left(2n-1\right)\left(2n-1-1\right)\left(2n-1+1\right)\)
\(=2n\left(2n-1\right)\left(2n-2\right)\)
=4n(n-1)(2n-1)
Vì n;n-1 là hai số nguyên liên tiếp
nên \(n\left(n-1\right)⋮2\)
=>\(4n\left(n-1\right)⋮4\cdot2=8\)
=>\(4n\left(n-1\right)\left(2n-1\right)⋮8\)
=>\(\left(2n-1\right)^3-\left(2n-1\right)⋮8\)
`a)55^{n+1}-55^n`
`=55^n(55-1)=54.55^n\vdots 54AAn\inZZ`
`b)n^2(n+1)+2n(n+1)`
`=(n+1)(n^2+2n)`
`=n(n+1)(n+2)`
Vì `n(n+1)(n+2)` là tích 3 số nguyên liên tiếp
`=>n(n+1)(n+2)\vdots 6AAn\inZZ`
`c)(2n-1)^3-(2n-1)`
`=(2n-1)[(2n-1)^2-1]`
`=(2n-1)(2n-1-1)(2n-1+1)`
`=2n(2n-2)(2n-1)`
`=4n(n-1)(2n-1)`
Vì `n(n-1)` là tích 2 số nguyên liên tiếp
`=>n(n-1)\vdots 2AAn\inZZ`
`=>4n(n-1)(2n-1)\vdots 8AAn\inZZ`
`=>dpcm`
