3
$\begingroup$

Does it exists a closed form (also approximating) for the following binomial weighted series?

$ \sum_{k=1}^n {n \choose k} \cdot k $

  • 0
    I just write an extra answer if you want a method without derivatives.2012-05-17

2 Answers 2

0

Without calculus answer:

Is easy to prove the next Lemma,

Lemma: $\boxed{k\dbinom{n}{k}=n\dbinom{n-1}{k-1}}.$

Therefore by Lemma,

$\sum_{k=1}^{n}k\dbinom{n}{k}=n\sum_{k=1}^{n}\dbinom{n-1}{k-1}.$

Then by binomial Newton's expression,

$\sum_{k=1}^{n}k\dbinom{n}{k}=n\sum_{k=0}^{n-1}\dbinom{n-1}{k}1^k1^{n-1-k}=n2^{n-1}\square$

3

By the binomial theorem: $ (x + 1)^n = \sum_{k=0}^{n} {n \choose k} x^k $

Differentiate both sides with respect to $x$:

$ n (x + 1)^{n-1} = \sum_{k=1}^{n} {n \choose k} k x^{k-1} $

Plug in $x = 1$:

$ n 2^{n-1} = \sum_{k=1}^{n} {n \choose k} k $

  • 0
    Many thanks Ayman.2012-05-16