Question

Im currently studying the running times of algorithms Big Oh notation. heres the question and my attempts.

Algorithms:
a) int sum = 0;
for(int i = 1; i < N; i < N; i *= 2)
for(int i = 0; i < n; i++)
sum++;

b) int sum = 0
for(int i = 1; i < N; i *=2)
for(int j = 0; j < N; j++)
sum++;

my answers:
a) (N-1) + (N-2) + (N-4) + (N-8) + ... + 1
then f(N) <= 4N-14
which is f(N) = O(N)??

b)N*(N-1) + N(N-2)+ N(N-4) + N(N-8) + ... + 1
then, N[(N-1) + (N-2) + (N-4) + (N-8) + ... + 1]
then f(N) = O(N^2)?