Asymptotic equivalent of the recurrence T(n)=3⋅T(n/2)+n
You have a mistake going from (2) to (3) because you lost $n$ and it is not clear why your sum should range from $i = 1$ to $i = n$ (it does not). Then it is not clear how you got from (3) to (4) where $n$ reappeared. Perhaps you just los $n$ in (3) but otherwise you had it on paper.

Let us try again:
\begin{align*}
T(n) &amp;= n + (3/2) n + (3/2)^2 n + \cdots
\end{align*}
But when does sum end? (This is what created your mistake.) It ends when we get to the base case, which is $T(1)$. After $k$ unfoldings we have $T(n/2^k)$ so we will reach $T(1)$ when $k = \log_2 n$. Thus, we continue like this:
\begin{align*}
T(n) &amp;= n + (3/2) n + (3/2)^2 n + \cdots + (3/2)^{\log_2 n} T(1) \\
&amp;= \textstyle \sum_{i=0}^{\log_2 n} (3/2)^i n\\
&amp;= n \cdot \textstyle \sum_{i=0}^{\log_2 n} (3/2)^i.
\end{align*}
It remains to add the sum. It is a geometric series, so we can look up the formula $\sum_{i=0}^k a^i = (a^{k+1} - 1)/(a - 1)$ and get
\begin{align*}
{\textstyle \sum_{i=0}^{\log_2 n} (3/2)^i} &amp;=
\frac{(3/2)^{1 +\log_2 n} - 1}{3/2 - 1} \\
&amp;= 2 \cdot ((3/2) \cdot (3/2)^{\log_2 n} - 1) \\
&amp;= 3 \cdot (3/2)^{\log_2 n} - 2 \\
&amp;= 3 \cdot (n^{\log_2 (3/2)}) - 2 \\
&amp;\in \Theta(n^{\log_2 (3/2)}) \approx \Theta(n^{0.58})
\end{align*}
The final result is this sum multiplied with $n$, so it is
$$\Theta(n \cdot n^{\log_2(3/2)}) = \Theta(n^{\log_2 2 + \log_2 (3/2)}) = \Theta(n^{\log_2 3}),$$
assuming I can still do logarithms....