We can apply the extended Masters Theorem to this problem
Comparing the given equation with T(n) = aT($\frac{n}{b}$) + nkLogp(n)
a= 2n, b=2, k=n, p=0 , we get a= bk . Hence the solution to recurrence will be
T(n)= Ɵ (nLogba.logp+1n) which is nnLog n.