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anonymous
 4 years ago
Great problem:
anonymous
 4 years ago
Great problem:

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Prove that \[\sum_{i=1}^{n}\sqrt{C_i}\le2^{n1}+\frac{n1}{2}\]

perl
 4 years ago
Best ResponseYou've already chosen the best response.0shoudlnt you start at i=0 ,or you intentionally left that out

nikvist
 4 years ago
Best ResponseYou've already chosen the best response.2\[\sqrt{C_i}\leq\frac{1}{2}(C_i+1)\]\[\sum\limits_{i=1}^{n}\sqrt{C_i}\leq\frac{1}{2}\sum\limits_{i=1}^{n}(C_i+1)\]\[\sum\limits_{i=1}^{n}\sqrt{C_i}\leq\frac{1}{2}\sum\limits_{i=1}^{n}C_i+\frac{n}{2}\]\[\sum\limits_{i=1}^{n}\sqrt{C_i}\leq\frac{1}{2}\left(\sum\limits_{i=0}^{n}C_i1\right)+\frac{n}{2}\]\[\sum\limits_{i=1}^{n}\sqrt{C_i}\leq\frac{1}{2}\left(2^n1\right)+\frac{n}{2}\]\[\sum\limits_{i=1}^{n}\sqrt{C_i}\leq2^{n1}+\frac{n1}{2}\]
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