Optimal choice of the sum

Question

I want the next sum to be maximal: $$p_{0,n-1}2^n:=p_0\ldots p_{n-1}2^n,\quad\sum_jq_j=1,\quad\sum_ip_i=1,$$ where $q_j$, and $p_i$ are probabilities, and $p_i\geq p_{i+1}$ for all $i$. The sum looks like this: $$\sum_{n=1}^{N}q_n\sum_{l=0}^{n-1}2^lp_{0,l-1}(1-p_l).$$ I calculated for $N = 2$, I got the following: $$q_1(1-p_0)+q_2[(1-p_0)+2p_0(1-p_1)]=1+p_0(2q_2(1-p_1)-1).$$ If $2q_2(1-p_1)-1$ is the maximum, then $p_0=1$. If $p_1=0$, then the optimal strategy depends on $2q_2-1>0$, so $(p_0,p_1)=(1,0)$, and if $2q_2-1<0$, so $p_0=0$. Unfortunately I didn't get along with the general case, could anyone help me? Thanks in advance!!!