For the case you mentioned(case (i)) you can use Arithmetic Progressions
In this case as first term=2*n-1 and the last term is n, so sum of all the terms is
S=n/2*(n+2*n-1)=O(n^2)
For the Case II, first term=(n+1-1)=n and last term=(n+n-1)=(2*n-1), so Sum, S is equal to,
S=n/2(first term+last term)=n/2*(n+2*n-1)=O(n^2)
You must have observed that S for both (i) & (ii) are same.