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Consider the following summation:

$$\sum_{k=0}^n k\binom{n}{k}(1-p)^{n-k}p^k$$

If the $k$ term was not present it would be a simple binomial. However, because of the $k$ term I am unable to derive the sum.

I believe the summation should be equal to $np$ (The summation is part of a larger expression, which upon solving using a different approach gives the $np$ result). I have tested this for smaller values of $n = \{1,2,3\}$.

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2 Answers 2

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Recall that for $k\geq 1$, $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ and therefore the sum can be evaluated as a simple binomial: $$\begin{align} \sum_{k=0}^n k\binom{n}{k}(1-p)^{n-k}p^k &=np\sum_{k=1}^n \binom{n-1}{k-1}(1-p)^{n-1-(k-1)}p^{k-1}\\ &=np((1-p)+p)^{n-1}=np. \end{align}$$ Along the same lines we show the more general identity: $$\sum_{k=0}^n k(k-1)\cdots (k-j)\binom{n}{k}(1-p)^{n-k}p^k =n(n-1)\cdots (n-j)p^{j+1}.$$

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  • $\begingroup$ @Samik It is a simple binomial! $\endgroup$
    – Robert Z
    Commented Mar 10, 2020 at 16:52
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Your expression is the mean of a binomial random variable with parameters $n$ and $p$. This is well known to be $np$.

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