Question

CMR:\(n^8+4n^7+6n^6+4n^5+n^4\)⋮16 với n∈Z

Answer 1

\(n^8+4n^7+6n^6+4n^5+n^4=n^4\left(n^4+4n^3+6n^2+4n+1\right)=n^4\left(n+1\right)\left(n^3+3n^2+3n+1\right)=n^4\left(n+1\right)\left(n+1\right)^3=n^4\left(n+1\right)^4=\left[n\left(n+1\right)\right]^4\)

Ta có \(n\left(n+1\right)\) là tích 2 số nguyên liên tiếp\(\Rightarrow n\left(n+1\right)⋮2\)\(\Rightarrow\left[n\left(n+1\right)\right]^4⋮16\)

Vậy \(n^8+4n^7+6n^6+4n^5+n^4⋮16\)