i goes 1 to n
j goes i to n
k goes j+1 to n
let's take first iteration for i: as i=1
j goes 1 to n
for j=1 , k: 2 to n = (n-1) times
j=2 , k: 3 to n = (n-2) times
...
j=n-1 , k: n to n = 1 time
j= n , k: 0 times
so, for i=1,
number of multiplications : 1 + 2 + ... + (n-1)
similarly for i = 2,
1+2+ .... + (n-2)
and so on,
for i = (n-1)
1
for i = n
0
so, for every i say last term as T
we have for every i, 1+2+3+... + T = T(T+1)/2 = (T^2 + T)/2
where T goes from 1 to (n-1)
∑ (T^2 + T)/2 , T goes 1 to (n-1) = (n-1)(n)(n+1)/ 6
so, option C is correct