0
$\begingroup$

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

1 Answers 1

1

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).

  • 0
    Maybe you mean $p_i=\dfrac{2^k}{2^{k+1}k-2^k+1}$ for every $0\leqslant i\leqslant k$ and $p_{k+1}=\dfrac{2^k(k-1)+1}{2^{k+1}k-2^k+1}$. Otherwise it doesn't sum to 1. By the way it doesn't work, for example for k=3.2012-05-28
  • 0
    I mean what is written, the $p_i$ do sum to $1$ and the example does work for every $k$. By the way I admire the way you approach the solutions suggested to you.2012-05-28
  • 0
    My mistake. Your solution is right. Thank you!2012-05-28