Where $T(1)=1$ and assuming that $n=2^k$ and that $k\ge 0$ and that the Master Theorm can't be used.
What I tried:
$T(n) = 2T(n/2)+n^2$
Following backwards substitution to get a pattern:
$T(2^k)=2T(2^{k-1}) +2^{2k} $
$=2T(2^{k-2}) + 2^{2k-2} + 2^{2k} $
$=2T(2^{k-3}) + 2^{2k-4} + 2^{2k-2} + 2^{2k} $
At this point I'm stuck as how to solve the rest of the problem - I think I'm suppose to get either the generalized $k$ form of the equation, and then either substitute in the $T(1)=1$ base case to find a generalized closed form solution, or solving it as a generic geometric progression, or substitute in a known summation formula but I'm not really sure what direction to take.
Starting with $T(2^k)=2T(2^{k-1})+2^{2k}$. Divide $2^k$ on both sides, we have $$\frac{T(2^k)}{2^k}=\frac{T(2^{k-1})}{2^{k-1}}+2^{k}$$ Let $S(k)=\frac{T(2^k)}{2^k}$, we have $S(1)=\frac{T(2)}2=\frac{2T(1)+2^2}2=3$ and $$S(k)=S(k-1)+2^k$$ $$\Rightarrow S(k)-S(k-1)=2^k$$ $$\Rightarrow S(k)=\sum_{r=2}^k[S(r)-S(r-1)]+S(1)=\sum_{r=2}^k2^r+3=4(2^{k-1}-1)+3=2^{k+1}-1$$ $$\Rightarrow \frac{T(2^k)}{2^k}=2^{k+1}-1$$ $$\Rightarrow T(2^k)=2^k(2^{k+1}-1)$$ $$\Rightarrow T(n)=n(2n-1)$$