I am wondering what the final form of:
$$ \sum_{j=0}^{N} {N\choose j} \left(\frac{c}{N}\right)^j\left(\frac{N-c}{N}\right)^{N-j} $$ where $c\in \mathbb{N}$ here is a constant smaller than $N$.
would be. Is there an easy way to do this? thanks
How to find the final form of $\sum_{j=0}^{N} {N\choose j} \left(\frac{c}{N}\right)^j\left(\frac{N-c}{N}\right)^{N-j}$?
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real-analysis
discrete-mathematics
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1 Answers
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This is an instance of the binomial theorem:
$$\sum_{j=0}^N\binom{N}j\left(\frac{c}N\right)^j\left(\frac{N-c}N\right)^{N-j}=\sum_{j=0}^N\binom{N}j\left(\frac{c}N\right)^j\left(1-\frac{c}N\right)^{N-j}=\left(\frac{c}N+\left(1-\frac{c}N\right)\right)^N=1\;.$$
Added in response to a comment: We can use the identity $j\binom{N}j=N\binom{N-1}{j-1}$ to write
$$\begin{align*} \sum_{j=0}^Nj\binom{N}j\left(\frac{c}N\right)^j\left(\frac{N-c}N\right)^{N-j}&=N\sum_{j=0}^N\binom{N-1}{j-1}\left(\frac{c}N\right)^j\left(\frac{N-c}N\right)^{N-j}\\ &=N\sum_{j=0}^{N-1}\binom{N-1}j\left(\frac{c}N\right)^{j+1}\left(\frac{N-c}N\right)^{N-1-j}\\ &=N\cdot\frac{c}N\sum_{j=0}^{N-1}\binom{N-1}j\left(\frac{c}N\right)^j\left(\frac{N-c}N\right)^{N-1-j}\\ &=c\left(\frac{c}N+\frac{N-c}N\right)^N\\ &=c\;. \end{align*}$$
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0If instead we have $\sum_{j=0}^Nj \binom{N}j\left(\frac{c}N\right)^j\left(\frac{N-c}N\right)^{N-j}$, is there an easy way to get the form from what you have? Thanks! – 2017-01-08
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0@user321627: It takes a little more work, but we can still get a simple closed form. – 2017-01-08