Zugdichte2r

2022-01-24

Find the value of $\sum _{k=1}^{n}k\left(\begin{array}{c}n\\ k\end{array}\right)$ ?

Damian Roberts

Beginner2022-01-25Added 14 answers

From the binomial theorem, we have

$(1+x)}^{n}=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{k$ (1)

Differentiating (1) reveals

$n{(1+x)}^{n-1}=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)k{x}^{k-1}$ (2)

Setting$x=1$ in (2) yields

$n{2}^{n-1}=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)k$

And we are done!

Differentiating (1) reveals

Setting

And we are done!

Deegan Mullen

Beginner2022-01-26Added 12 answers

We have

$\sum _{k=0}^{n}k\left(\begin{array}{c}n\\ k\end{array}\right)=\sum _{k=1}^{n}k\left(\begin{array}{c}n\\ k\end{array}\right)$

$n=\sum _{k=1}^{n}\frac{(n-1)!}{(k-1)!(n-k)!}$

$n\sum _{l=0}^{n-1}\frac{(n-1)!}{l((n-1)-l)!}$

$=n\sum _{l=0}^{n-1}\left(\begin{array}{c}n-1\\ l\end{array}\right)$

$=n{2}^{n-1}$

RizerMix

Expert2022-01-27Added 656 answers

Proof without derivatives:
$\sum _{k=1}^{n}k((n),(k))=\sum _{k=1}^{n}k\frac{n!}{(n-k)!k!}$
$=\sum _{k=1}^{n}\frac{n!}{(n-k)!(k-1)!}$
$=\sum _{k=1}^{n}\frac{n(n-1)!}{(n-k)!(k-1)!}$
$=\sum _{k=1}^{n}n((n-1),(k-1))$
$=n\sum _{k=0}^{n-1}((n-1),(k))=n{2}^{n-1}$
Alternate proof via probability theory:
Toss a fair coin n times, find the expected no of heads. Let N be the random variable denoting the number of heads. Then $E[N]=n/2$ because N is the sum of n bernoulli random variables with probability 1/2. But we also know that N has a binomial distribution. Hence
$E[N]=\sum _{k=1}^{n}k((n),(k)){2}^{-n}$
Rearrange to get your answer.

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