$T(n+1)=T(n)+\lfloor \sqrt{n+1} \rfloor$ $\forall n\geq 1$
$T(1)=1$
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:
$T(n)=T(n-1)+\sqrt{n}$
$T(n-1)=T(n-2)+\sqrt{n-1}$
therefore,
$T(n)=T(n-2)+\sqrt{n-1}+\sqrt{n}$
$T(n)=T(n-3)+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}$
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$T(n)=T(n-(n-1))+...T(n-k)+\sqrt{n-(k-1)}+...+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}$
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$T(n)=T(1)+...T(n-k)+\sqrt{n-(k-1)}+...+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}$
$T(n)=T(1)+...+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}$
I'm stuck up here, how to solve this ahead?
Noting that $T(1)=1=\lfloor \sqrt{1}\rfloor$, so far, you've narrowed it down to the form (although you forgot the floor symbols, which are important):
$$T(n)=\sum_{i=1}^{n}\lfloor\sqrt{n}\rfloor$$
which is correct; to solve this, notice that if we calculated $T(4^2-1)=T(15)$ thusly, for instance, the sum would expand as (after we lay it out very suggestively): $$\begin{align*}T(n)&=1+1+1\\&+2+2+2+2+2\\&+3+3+3+3+3+3+3\end{align*}$$ Notice the pattern here: The is $3$ integers such that $\lfloor \sqrt{n}\rfloor =1$ - that is $i$ with $1^2\leq i < 2^2$. There are $5$ such that $\lfloor \sqrt{n}\rfloor =3$ - that is, those with $2^2 \leq i < 3^2$. More generally, there are $2n+1$ integers $i$ such that $\lfloor \sqrt{i}\rfloor =n$, being those in the range $[n^2,(n+1)^2)$ which has length $$(n+1)^2-n^2=2n+1.$$
To use this, consider $T(n^2-1)$. The sum will look like: $$\begin{align*}T(n)&=1+1+1\\&+2+2+2+2+2\\&+3+3+3+3+3+3+3\\&\qquad\qquad \vdots\\&+\underbrace{(n-1)+(n-1)+\ldots+(n-1)}_{\text{$2n-1$ times}}\end{align*}$$ which can be grouped as: $$\begin{align*}T(n)&=3\cdot 1\\&+5\cdot 2\\&+7\cdot 3\\&\quad\,\,\vdots\\ &+(2i+1)\cdot i\\&\quad\,\,\vdots \\&+(2n-1)\cdot (n-1)\end{align*}$$ or, as a sum: $$T(n^2-1)=\sum_{i=1}^{n-1}(2i+1)\cdot i = \sum_{i=1}^{n-1}2i^2 + i$$ and knowing the identities that $\sum_{i=1}^{n-1}i = \frac{1}2(n^2-n)$ and $\sum_{i=1}^{n-1}i^2=\frac{1}6(2n^3-3n^2+n)$ which can easily be verified (or found) using induction, the above can be written as $$T(n^2-1)=\frac{2}3n^3 -\frac{1}2n^2-\frac{1}6n.$$ From here, any other value is easy to determine - for instance, if $x$ is such that $x<(n+1)^2$ and $x=(n^2-1)+c$, then $T(x)=T(n^2-1)+cn$.