If $T(n+1)=T(n)+\lfloor \sqrt{n+1} \rfloor$ $\forall n\geq 1$, what is $T(m^2)$?

$T(n+1)=T(n)+\lfloor \sqrt{n+1} \rfloor$ $\forall n\geq 1$

$T(1)=1$

The value of $T(m^2)$ for m ≥ 1 is?


Clearly you cannot apply master theorem because it is not of the form $T(n)=aT(\frac{n}{b})+f(n)$

So I tried Back Substitution:

$T(n)=T(n-1)+\lfloor\sqrt{n}\rfloor$

$T(n-1)=T(n-2)+\lfloor\sqrt{n-1}\rfloor$

therefore,

$T(n)=T(n-2)+\lfloor\sqrt{n-1}\rfloor+\lfloor\sqrt{n}\rfloor$

$T(n)=T(n-3)+\lfloor\sqrt{n-2}\rfloor+\lfloor\sqrt{n-1}\rfloor+\lfloor\sqrt{n}\rfloor$

.

.

$T(n)=T(n-(n-1))+...T(n-k)+\lfloor\sqrt{n-(k-1)}\rfloor+...+\lfloor\sqrt{n-2}\rfloor+\lfloor\sqrt{n-1}\rfloor+\lfloor\sqrt{n}\rfloor$

.

.

$T(n)=T(1)+...T(n-k)+\lfloor\sqrt{n-(k-1)}\rfloor+...+\lfloor\sqrt{n-2}\rfloor+\lfloor\sqrt{n-1}\rfloor+\lfloor\sqrt{n}\rfloor$

$T(n)=T(1)+...+\lfloor\sqrt{n-2}\rfloor+\lfloor\sqrt{n-1}\rfloor+\lfloor\sqrt{n}\rfloor$

I'm stuck up here and the answer is given as -


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

how to solve and reach the answer?

For a slightly different way to look at his problem, consider your original equation,
$$
T(n) = T(1)+\lfloor\sqrt2\rfloor+\lfloor\sqrt3\rfloor+\dotsb+\lfloor\sqrt n\rfloor
$$
Now the key here is to group the terms with the same values of $\lfloor\sqrt k\rfloor$:
$$\begin{align}
T(n) &= (T(1)+\lfloor\sqrt2\rfloor+\lfloor\sqrt3\rfloor)\\
&+(\lfloor\sqrt4\rfloor+\lfloor\sqrt5\rfloor+\lfloor\sqrt6\rfloor+\lfloor\sqrt7\rfloor+\lfloor\sqrt8\rfloor)\\
&+(\lfloor\sqrt9\rfloor+\lfloor\sqrt{10}\rfloor+\lfloor\sqrt{11}\rfloor+\lfloor\sqrt{12}\rfloor+\lfloor\sqrt{13}\rfloor+\lfloor\sqrt{14}\rfloor+\lfloor\sqrt{15}\rfloor)\\
&+\dotsc
\end{align}$$
and observe that each summand will have $2k+1$ terms, since the difference $(k+1)^2-k^2 = 2k+1$:
$$
T(n) = (1\cdot3)+(2\cdot5)+(3\cdot 7)+(4\cdot 9)+\dotsb
$$
so we'll have, with $n=m^2$
$$\begin{align}
T(m^2) &= (1\cdot3)+(2\cdot5)+\dotsb+\left(\left\lfloor\sqrt{(m-1)^2}\right\rfloor+\dotsb+\left\lfloor\sqrt{m^2-1}\right\rfloor\right)+\left\lfloor\sqrt{m^2}\right\rfloor\\
&=\sum_{k=1}^{m-1}k(2k+1)+m =\sum_{k=1}^{m-1}(2k^2+k)+m\\
&=2\sum_{k=1}^{m-1}k^2+\sum_{k=1}^{m-1}k+m\\
&=2\frac{(m-1)(m)(2(m-1)+1)}{6}+\frac{(m-1)(m)}{2}+m\\
&=\frac{m}{6}(4m^2-3m+5)
\end{align}$$
as required.

StackExchange Computer Science Q#35558, answer score: 3