Q35

MCQ2 marks, −0.5

Consider the following recurrence relation $T(1)=1$ $T(n+1) = T(n)+\lfloor \sqrt{n+1} \rfloor$ for all $n \geq 1$ The value of $T(m^2)$ for $m \geq 1$ is

(A)

$\frac{m}{6}\left(21m-39\right)+4$

(B)

$\frac{m}{6}\left(4m^2-3m+5\right)$

(C)

$\frac{m}{2}\left(3m^{2.5}-11m+20\right)-5$

(D)

$\frac{m}{6}\left(5m^3-34m^2+137m-104\right)+\frac{5}{6}$

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