Use the loop invariant (I) to show that the code below correctly computes n P−1 k=0 2k
(this sum represents the sum of the first n even integers where n ≥ 1).
Algorithm 1 evenSum(int n)
1: p = 2(n − 1)
2: i = n − 1
3: while i > 0 do
4: //(I) p = nP−1 k=i 2k
5: i − −
6: p = p + 2i
7: end while
8: return p
Prove the Base case of the loop invariant (I):