How do I take out (n+1) from the recurrence a(n)=(n+1/n)a(n-1)+3n+3?

Answer 1

I'd divide by (n+1)

((n+1/n)a(n-1)+3n+3)/(n+1)=

n/(n+1)+1/(n^2+n)(an-a)+(3n+3)/(n+1)

=3

so; n/(n+1)+(an-a)/(n^2+n)+3=a(n)