first loop is executed for n times i=0,1....n-1
second loop is called for every value of i, so second loop runs for 0 iterations then 1 then 2, ...... and finally n-1
so in total 1+2+3+....+n-1 = (n-1)(n)/2 times
third loop runs for every value of j
so first when j=1, third loop runs once
when j=2, third loop runs for k=0 and k=1 so twice
And similarly when j= n-1 total n-1 loops
1+(1+2)+(1+2+3)+(1+2+3+4)+.....+(1+2+3+4....n-1)
=(n-1)(1) +(n-2)(2)+ (n-3)(3)+.......(1)(n-1)
=(n-1)(1) +(n-2)(2)+ (n-3)(3)+.......(n-(n-1))(n-1)
=[n+2n+3n+4n+5n+6n+.......(n-1)n]-[ 12+22+32+42+....(n-1)2]
=[n(1+2+3+4+5+6+.......(n-1)]-[ 12+22+32+42+....(n-1)2]
=n(n-1)(n)/2 - [12+22+32+42+....(n-1)2]
=[n(n-1)(n)/2] - [(n-1)*n*(2n-2+1)/6]
=(n3+n)/6
=$\Theta$(n3)