Need in detail solution of the problem below:

Given the following source code where n = number of data points:

for( int i = 0; i < 100*n; ++i) { // cost for line?

for( int j = 0; j < 50*n; ++j) { // cost for line?

sum += i*j; // cost for line?

}

}

It is found the cost function of the code is:

T(n) = c0*(100*n + 1) + c1*(100*n)*(50n + 1) + c2*100n*50n.

Reducing and collecting produces:

T(n) = c0*100n + c0 + c1*5000n^2 + c1*100n + c2*5000n^2

T(n) = c0 + c0*100n + c1*100n + c1*5000n^2 + c2*5000n^2

T(n) = (c0 + c0*100 + c1*100)n + (c1*5000 + c2*5000)n^2

Replacing the constant terms with the values c3 = c0 + c0*100 + c1*100 and c4 = c1*5000 + c2*5000.

Produces:

T(n) = c4*n² + c3*n

Determine which is true:

For any c3,c4 > 0 is it true T(n) (n²) as n-> .

For any c3,c4 > 0 is it true T(n) q(n²) as n-> .

For any c3,c4 > 0 is it true T(n) O(n²) as n-> .

Show all logic and work using the inequality proof tests:

i.e. for f(n) O(g(n)) show that f(n) <= c*g(n) where c > 0 as n-> .