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

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

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I just write an extra answer if you want a method without derivatives. –  Gastón Burrull May 17 '12 at 4:59

2 Answers 2

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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,

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

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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 $x = 1$:

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

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Many thanks Ayman. –  fede May 16 '12 at 12:39

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