computing the following sum with few calculations

Question

Let $p\in(0,1)$ and $n\in\mathbb{N}$. Is there a trick that computes the sum $$\sum_{k=0}^n k {n \choose k }p^k(1-p)^{n-k}$$ without much work and all? I think it is equal to $n\cdot p$.

Answer 1

The summation can be recognized as the expectation of a random variable $X$ having binomial distribution with parameters $n$ and $p$.

This random variable can be written as:$$X=B_1+\cdots+B_n$$where the $B_i$ have Bernoulli-distribution with parameter $p$.

Then with linearity of expectation we find:$$\mathbb EX=\mathbb EB_1+\cdots+\mathbb EB_n=p+\cdots+p=np$$

Answer 2

Hint: Use $$ k\binom{n}{k}=n\binom{n-1}{k-1}. $$ $$np\sum_{k=0}^n\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}=? $$

Or by differentiating the expression $$ \sum_{k=0}^n\binom{n}{k}a^kb^{n-k}=(a+b)^n $$ with respect to $a$, derive $$ \sum_{k=0}^nk\binom{n}{k}a^{k-1}b^{n-k}=n(a+b)^{n-1}. $$ Let $a=p$ and $b=1-p$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{k = 0}^{n}k{n \choose k }p^{k}\pars{1 - p}^{n - k}} = \pars{1 - p}^{n}\sum_{k = 0}^{n}{n \choose k }k\pars{p \over 1 - p}^{k} \\[5mm] = &\ \left.\pars{1 - p}^{n}\,\,\alpha\,\partiald{}{\alpha} \sum_{k = 0}^{n}{n \choose k }\alpha^{k}\,\right\vert_{\ \alpha\ =\ p/\pars{1 - p}} = \left.\pars{1 - p}^{n - 1}\,p\,\partiald{\pars{1 + \alpha}^{n}}{\alpha} \,\right\vert_{\ \alpha\ =\ p/\pars{1 - p}} \\[5mm] = &\ \pars{1 - p}^{n - 1}\,p\bracks{n\pars{1 + \alpha}^{n - 1}} _{\ \alpha\ =\ p/\pars{1 - p}} = \pars{1 - p}^{n - 1}\,pn\pars{1 + {p \over 1 - p}}^{n - 1} = \bbx{np} \end{align}