no need to solve it just see the leading term in s1 and s2 .
In s1 as the value of r is increased the terms keep on decreasing due to exponential function in denominator. So the term where r =1 I.e n/2 will be the leading and thus it will be the order of n.
In S2 the last term where r=logn-1 is the leading term as exponential function is present in numerator so if you simplify it you would get Nlogn.