Question

Great problem:

Answer 1

...what is?

Answer 2

Prove that
\[\sum_{i=1}^{n}\sqrt{C_i}\le2^{n-1}+\frac{n-1}{2}\]

Answer 3

what does Ci stand for?

Answer 4

Combination{n,i}

Answer 5

we know that sum Ci = 2^n

Answer 6

Yes we do

Answer 7

shoudlnt you start at i=0 ,or you intentionally left that out

Answer 8

Intentional

Answer 9

\[\sqrt{C_i}\leq\frac{1}{2}(C_i+1)\]\[\sum\limits_{i=1}^{n}\sqrt{C_i}\leq\frac{1}{2}\sum\limits_{i=1}^{n}(C_i+1)\]\[\sum\limits_{i=1}^{n}\sqrt{C_i}\leq\frac{1}{2}\sum\limits_{i=1}^{n}C_i+\frac{n}{2}\]\[\sum\limits_{i=1}^{n}\sqrt{C_i}\leq\frac{1}{2}\left(\sum\limits_{i=0}^{n}C_i-1\right)+\frac{n}{2}\]\[\sum\limits_{i=1}^{n}\sqrt{C_i}\leq\frac{1}{2}\left(2^n-1\right)+\frac{n}{2}\]\[\sum\limits_{i=1}^{n}\sqrt{C_i}\leq2^{n-1}+\frac{n-1}{2}\]