Consider the following recurrence relation The value of T(m2) for m ≥ 1 is

We have T(1) = 1 and T(n+1) = T(n) + floor(sqrt(n+1)). Therefore

T(m^2) = T(1) + sum_{i=2}^{m^2} floor(sqrt(i)) = 1 + (sum_{i=1}^{m^2} floor(sqrt(i)) - 1) = sum_{i=1}^{m^2} floor(sqrt(i)).

1. Group terms by the value of floor(sqrt(i)). For j = 1,2,...,m-1, floor(sqrt(i)) = j for i = j^2,...,(j+1)^2 - 1, which gives 2j+1 terms. For j = m only i = m^2 is included, giving 1 term.

2. Thus sum_{i=1}^{m^2} floor(sqrt(i)) = sum_{j=1}^{m-1} j(2j+1) + m.

3. Evaluate the sums: sum_{j=1}^{m-1} j(2j+1) = 2 sum_{j=1}^{m-1} j^2 + sum_{j=1}^{m-1} j.

4. Use closed forms: sum_{j=1}^{m-1} j = (m-1)m/2 and sum_{j=1}^{m-1} j^2 = (m-1)m(2m-1)/6. Substituting gives

5. sum_{i=1}^{m^2} floor(sqrt(i)) = 2 * (m-1)m(2m-1)/6 + (m-1)m/2 + m = (4m^3 - 3m^2 + 5m)/6.

6. Therefore T(m^2) = (4m^3 - 3m^2 + 5m)/6 = (m/6)(4m^2 - 3m + 5).

Final answer: T(m^2) = (m/6)(4m^2 - 3m + 5).