Stackelberg Game with three players

My friend and I were solving the following game for assignment, and came up with different solutions. Could any one help us know whose solution is right?

Suppose there are 3 firms in a homogeneous good market, which set their quantities sequentially (firm $$i$$ in period $$i$$). Suppose firms have zero marginal cost of production.The industry faces an inverse demand $$P(q) = 16-q$$ where $$q=\sum_{i=1}^3q_i$$ is the aggregate quantity. Solve for $$q_1$$.

Both agreed to the following:(See Rigorous solution of Stackelberg leader-follower game with N-players?) $$q_i = \frac{1}{2}(16-\sum_{j=1}^{i-1}q_j)$$ $$q_i = \frac{1}{2}q_{i-1}$$

My Solution: $$max_{q_1}Pq_1=max(16-q_1-\frac{1}{2}q_1-(\frac{1}{2})^2q_1)q_1$$ $$=(16-\frac{7}{4}q_1)q_1$$ Applying the FOC results in $$q_1=32/7$$ with the profit level $$\frac{32\cdot8}{7}$$.

Her Solution: Applying i=1 to the first equation provides $$q_1=8$$. Hence, the profit level is $$(16-(8+4+2))\times 8 = 16$$.

Could anyone tell us which solution is right?

Firm 3 will choose $$q_3$$ after observing $$q_1$$ and $$q_2$$ in the following way. It solves the following problem:

$$\begin{eqnarray*}\max_{q_3\geq 0} & \ (16-q_1-q_2-q_3)q_3\end{eqnarray*}$$

and we get $$q_3=\dfrac{16-q_1-q_2}{2}$$

Firm 2 will anticipate firm 3's response as above and choose $$q_2$$ after observing $$q_1$$ in the following way. It solves the following problem:

$$\begin{eqnarray*}\max_{q_2\geq 0} & \ (16-q_1-q_2-q_3)q_2 \\ \text{s.t.} & q_3=\dfrac{16-q_1-q_2}{2}\end{eqnarray*}$$

which can be re-written as $$\begin{eqnarray*}\max_{q_2\geq 0} & \ \frac{(16-q_1-q_2)q_2}{2} \end{eqnarray*}$$

and we get $$q_2=\dfrac{16-q_1}{2}$$.

Firm 1 will anticipate firm 2 and 3 to respond in the manner above and choose $$q_1$$ by solving $$\begin{eqnarray*}\max_{q_1\geq 0} & \ (16-q_1-q_2-q_3)q_1 \\ \text{s.t.} & q_3=\dfrac{16-q_1-q_2}{2} \\ \text{and} & q_2=\dfrac{16-q_1}{2}\end{eqnarray*}$$

which can be re-written as

$$\begin{eqnarray*}\max_{q_1\geq 0} & \ \frac{(16-q_1)q_1}{4} \end{eqnarray*}$$

and we get $$q_1=8$$.