Does it exists a closed form (also approximating) for the following binomial weighted series?
$ \sum_{k=1}^n {n \choose k} \cdot k $
Does it exists a closed form (also approximating) for the following binomial weighted series?
$ \sum_{k=1}^n {n \choose k} \cdot k $
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,
$\sum_{k=1}^{n}k\dbinom{n}{k}=n\sum_{k=0}^{n-1}\dbinom{n-1}{k}1^k1^{n-1-k}=n2^{n-1}\square$
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 $