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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

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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).

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    My mistake. Your solution is right. Thank you!2012-05-28