Given $k$, find $n \in N$ and $p_i$ such that
$$\sum_{i=0}^n p_i 2^i = 2^k$$ $$\sum_{i=0}^n p_i = 1$$ $$0
Given $k$, find $n \in N$ and $p_i$ such that
$$\sum_{i=0}^n p_i 2^i = 2^k$$ $$\sum_{i=0}^n p_i = 1$$ $$0
Try $n=k+1$, $p_i=\dfrac{2^k}{2^{k+1}k+1}$ for every $0\leqslant i\leqslant k$ and $p_{k+1}=\dfrac{2^k(k-1)+1}{2^{k+1}k+1}$ (knowing that there are plenty of other solutions).