Q2 double sumexponentialsint n n is a power of 3 ie n3k or k

Q2.
double sum_exponentials(int n)
//n is a power of 3, i.e., n=3k or k=log3n
{
int sum=0;
for (int i=1; i<n; i=i*3)
sum = sum + i;
return sum;
}

Solution

Explanation:
Time Complexity of a loop is considered as O(Logn) if the loop variables is divided / multiplied by a constant amount.

for (int i = 1; i <=n; i *= c) {
// some O(1) expressions
}
for (int i = n; i > 0; i /= c) {
// some O(1) expressions
}

So, your function is:

double sum_exponentials(int n)
//n is a power of 3, i.e., n=3k or k=log3n
{
   int sum=0; // O(1) time to run this line
   for (int i=1; i<n; i=i*3) //it iterates logn times where base of log function is 3
       sum = sum + i; // this takes O(1) times , logn (base 3) times it executes
   return sum; // O(1)
}

Total : O(1) + logn (base 3) + O(1)
= O(logn)