$T(m^2) = T(m^2-1) + \left\lfloor\sqrt{(m^2)} \right\rfloor$
$= T(m^2 - 2) + \left\lfloor\sqrt{(m^2 - 1)} \right\rfloor +\left\lfloor\sqrt{(m^2)} \right\rfloor$
$= T(m^2 - 3) + \left\lfloor\sqrt{(m^2 - 2)} \right\rfloor + \left\lfloor\sqrt{(m^2 - 1)} \right\rfloor +\left\lfloor\sqrt{(m^2)} \right\rfloor$
$\vdots$
$= T(1) + \left\lfloor\sqrt{(2)} \right\rfloor + \left\lfloor\sqrt{(3)} \right\rfloor + \ldots + \left\lfloor\sqrt{(m^2)} \right\rfloor$
$= 3 \times 1 + 5 \times 2 + \ldots + \left(2m - 1\right) \times (m-1) +m $
(We are taking floor of square root of numbers, and between successive square roots number of numbers are in the series $3,5,7 \dots$ like $3$ numbers from $1..4$, $5$ numbers from $5-9$ and so on).
We can try out options here or solve as shown at end:
Put $m = 5$, $T(25) = 3 \times 1 + 5 \times 2 + 7 \times 3 + 9 \times 4 + 5 = 75$
- $59$
- $75$
- non-integer
- $297.5$
So, answer must be B.
$T(m^2) = 3 \times 1 + 5 \times 2 + \dots + \left(2m - 1\right) \times (m-1) +m$
$= m + \displaystyle \sum_{i=1}^{m-1} \left[ (2i+1). (i) \right] $
$= m + \displaystyle \sum_{i=1}^{m-1} \left[2i^2 + i\right]$
$= m + \frac{(m-1) .m .(2m-1)}{3} + \frac{(m-1)m}{2}$
$= \frac{m}{6} \left(6 + 4m^2 -2m -4m + 2 + 3m - 3\right)$
$= \frac{m}{6} \left(4m^2 -3m + 5\right) $
- Sum of the first $n$ natural numbers $=\frac{n. (n+1)}{2}.$
- Sum of the squares of first $n$ natural numbers $ = \frac{n. (n+1). (2n+1)}{6}.$