Show (n+a)^b=Ɵ(n^b), for any real constants a and b where b>0

Here, using binomial theorem for expanding (n+a)^b, we get,
C(b,0)n^b+C(b,1)n^(b-1)a+.....+C(b,b-1)na^(b-1)+C(b,b)a^b
We can obtain some constants such that (n+a)^b<=c1*(n^b), for all n>=n0
and (n+a)^b >=c2*(n^b), for all n>=n0

Here we may take c1=2^b, c2=1, n0=|a|
Since, 1*(n^b)<=(n+a)^b<=2^b*(nb)

Hence, the problem is solved.