Best solutions to - "Find the value of ∑k=1nk(nk)?"

Zugdichte2r

Answered question

2022-01-24

Find the value of

\sum_{k = 1}^{n} k (\begin{matrix} n \\ k \end{matrix})

Answer & Explanation

Damian Roberts

Beginner2022-01-25Added 14 answers

From the binomial theorem, we have

{(1 + x)}^{n} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) x^{k}

(1)
Differentiating (1) reveals

n {(1 + x)}^{n - 1} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) k x^{k - 1}

(2)
Setting

x = 1

in (2) yields

n 2^{n - 1} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) k

And we are done!

Deegan Mullen

Beginner2022-01-26Added 12 answers

We have

\sum_{k = 0}^{n} k (\begin{matrix} n \\ k \end{matrix}) = \sum_{k = 1}^{n} k (\begin{matrix} n \\ k \end{matrix})

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} (\begin{matrix} n - 1 \\ l \end{matrix})

= 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.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?