My interpretation of your question, my hand calculations and explanations are as follows. For periods of six months you have:
At the beginning of period 1 you invest $C_{0}=1000$ (dollars).
At the end of period 1 $C_{0}$ worths $F_{1}=1200$.
At the beginning of period 2 you invest the additional capital of $%
C_{1}=2000$.
At the end of period 2 your investement worths $F_{2}=4000$.
Figure: Cash Flows of -1000, -2000 and +4000 dollars
Let us denote by $r$ the anual nominal interest rate of your investement. In
$n$ periods of six months the investement worths $(1+r/2)^{n}$ per currency unit of invested capital.
The initial investement $C_{0}$ worths $C_{0}(1+r/2)^{2}$ at the end of
period 2.
The additional capital $C_{1}$ worths $C_{1}(1+r/2)$ at the end of period 2.
Then we have
$C_{0}(1+r/2)^{2}+C_{1}(1+r/2)=F_{2}$
$1000(1+r/2)^{2}+2000(1+r/2)=4000$
or
$(1+r/2)^{2}+2(1+r/2)=4$
$2r+\dfrac{1}{4}r^{2}+3=4.$
The solution is: $r=2\sqrt{5}-4\approx 0.47214\approx 47.214\%$ anual
nominal rate
To determine the effective interest rate, you could find how much would you
need to invest so that at a nominal rate of $r$ you yould have $4000$ in 1
year (2 periods):
$P\cdot 1.2361^{2}=4000$
$P=4000/1.2361^{2}=2617.9$
The anual effective interest rate is
$i_{eff}=\dfrac{4000}{2617.9}-1=0.52794\approx 52,794\%$.
Remark: The future value $F_{n}$ at the end of period $n$ of a present value
$P$ is given by
$F_{n}=P\left( 1+\dfrac{r}{m}\right) ^{n}$,
where $r=i_N$ is the anual nominal interest rate compounded $m$ periods per year. In your case $m=2$.
Reference: Rigg, Bedworth and Randhawa, Engineering Economics, Mc Graw Hill, 4th ed., 1996.
All errors and omissions are of mine, of course.