Question

CMR với mọi số tự nhiên n thì \(A=\)\(5^{n+2}+26.5^n+8^{2n+1}⋮59\)

Answer 1

\(A=5^{n+2}+26.5^n+8^{2n+1}\left(n\in N\right)\)

\(=25.5^n+26.5^n+8.64^n\)

\(=5^n\left(25+26\right)+8.64^n\)

\(=5^n\left(59-8\right)+8.64^n\)

\(=59.5^n+8\left(64^n-5^n\right)\)

\(=59.5^n+8\left(64-5\right)\left(64^{n-1}+64^{n-2}.5+...\right)\)

\(=59.5^n+8.59\left(64^{n-1}+64^{n-2}.5+...\right)\)

\(=59\left[5^n+8\left(64^{n-1}+64^{n-2}.5+...\right)\right]⋮59\)

Vậy \(A⋮59\)\(\forall n\in N\)(đpcm)