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Let $C_k^i=\frac{k!}{i!(k-i)!}$. Show that $$\lim_{k\to\infty}\frac{C_k^i+C_k^{n+i}+\cdots+C^{([k/n]-1)n+i}_k}{2^k}=\frac{1}{n}$$ for any $1\leq i<n$. Here $i,n$ be positive integers.

As is well-known, $\sum_{i=0}^k C_k^i=2^k$. But how to prove the above limit? Choose $C_k^i$ after $n$ blocks.

xldd
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    Are you sure you have the limit of summation correct here? Supposing $n | k$, what does $C_{k}^{k+i}$ mean to you? – Chris Jan 07 '18 at 05:26

2 Answers2

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Suppose that we have $k$ switches, each one can either be on or off. There are $n$ lights, numbered $0,1,\dots,n-1$. The $i$-th light will be on if and only if the number of switches which are on is congruent to $i$ modulo $n$.

The probability that the $i$-th light is on is $\displaystyle \frac{C_k^i+C_k^{n+i}+\cdots+C^{[k/n]n+i}_k}{2^k}$.

There is exactly one of the lights being on. As $k\to \infty$, each light has equal chance to be on. So the probability that a particular light is on is $\displaystyle \frac{1}{n}$.

CY Aries
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This should help.

$$ \lim_{k \rightarrow \infty}\frac{C^i_k+C^{n+1}_k+...}{2^k}=\lim_{k \rightarrow \infty}\frac{\frac{1}{n}\sum_{j=0}^{n-1}(2 \cos{\frac{\pi j}{n}})^k \cos{\frac{\pi kj}{n}}}{2^k}$$ and because $k \rightarrow \infty$, only first term in sum is different from 0 therefore, $$\lim_{k \rightarrow \infty}\frac{C^i_k+C^{n+1}_k+...}{2^k} = \frac{1}{n}$$.