Q-2) Given the below: (i) 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) (n2) as n-> . For any c3,c4 > 0 is it true T(n) (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-> .

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