f(n)=n^3+5nf(n+1)=(n+1)^3+5(n+1)=n^3+3n^2+3n+1+5n+5=(n^3+5n)+3n^2+3n+6=f(n)+3(n^2+n+2)=f(n)+3[(n+1)n+2]因为n∈R,则f(1)=1+5=6 能被2,3,6整除,所以只用证明f(n+1)=能被2,3,6整除.f(n+1)=f(n)+3[(n+1)n+2],很明...