Solution to recurrence $T(n) = T(n/2) + n^2$
Try this way:
$$
\begin{align*}
T(n) &amp;= T(n/2) + n^2 \\
T(n) &amp;= T(n/4) + (n/2)^2 + n^2\text{ (expand $T(n/2)$)} \\
T(n) &amp;= T(n/8) + (n/4)^2 + (n/2)^2 + n^2\text{ (expand $T(n/4)$)}
\end{align*}
$$

You will end up with:
$$
\begin{align*}
T(n) &amp;= n^2 + (n/2)^2 + (n/4)^2 + (n/8)^2 + \cdots + (n/2^{\lg n})^2 \\
T(n) &amp;= n^2 + n^2/4 + n^2/16 + n^2/64 + \cdots
\end{align*}
$$

We can bound this from above as $T(n) \le 2 n^2$ and from below with $T(n) \ge n^2$, giving $T(n) = \Theta(n^2)$.

Note: we do not need to use induction to prove this; since we can create an upper bound of $T(n) \le c_1f(n)$ and a lower bound of $T(n) \ge c_2f(n)$, we can use the definition of $\Theta(n)$ to establish the time complexity of the relation.