Summation of ${\sum_{k=1}^n k\binom{n}{k} = n2^{n-1}}$ using equations

Question

I was trying to solve the summation:

$${\sum_{k=1}^n k\binom{n}{k} = n2^{n-1}}$$

I started something like:

$${\sum_{k=0}^n \binom{n}{k} = 2^n}$$ $${\Rightarrow \sum_{k=1}^n \binom{n}{k} = 2^n - 2^0 \qquad(1)}$$ Similarly, $${\Rightarrow \sum_{k=2}^n \binom{n}{k} = 2^n - 2^1 \qquad(2)}$$ $${\vdots}$$ $${\Rightarrow \sum_{k=n}^n \binom{n}{k} = 2^n - 2^{n-1} \qquad(n)}$$

On adding the above n equations, we get, $${\sum_{k=1}^n k\binom{n}{k} = n2^n - (2^0 + 2^1 + \cdots + 2^{n-1})}$$ As the expression inside brackets on RHS is a GP, the sum comes up as, $${S_n = 2^0\left(\frac{2^n - 1}{2-1}\right) = 2^n - 1}$$ $${\Rightarrow \sum_{k=1}^n k\binom{n}{k} = n2^n - 2^n + 1 \neq n2^{n-1}}$$

I've checked my solution many times and have searched for similar solutions here and here, but there was no answer using this method. What I am doing wrong?

I think you're missing terms. I think you should have $2^n - 2^1 - 2^0$, for instance in equation 2. — Cameron L. Williams, Apr 24 '19 at 04:09
It would not occur to me not to use the differentiation trick in a highly voted answer to the linked question. Anyway, something's wrong in your equations $(2)$ thru $(n)$. Surely $$\sum_{k=2}^n\binom n k=2^n-\binom n 0-\binom n1=2^n-1-n$$ and $$\sum_{k=n}^n\binom nk=1.$$ — Jyrki Lahtonen, Apr 24 '19 at 04:24

score 5 · Answer 1 · answered Apr 24 '19 at 05:24

\begin{align} &\bbox[10px,#ffd]{\sum_{k = 1}^{n}k{n \choose k}} = \bbox[10px,#fee]{\sum_{k = 0}^{n}k{n \choose k}} = \sum_{k = 0}^{n}\left(n - k\right){n \choose n - k} \\[5mm] = &\ n\sum_{k = 0}^{n}{n \choose k} - \bbox[10px,#fee]{\sum_{k = 0}^{n}k{n \choose k}} \\[5mm] &\ \implies \bbox[10px,#fee]{\sum_{k = 0}^{n}k{n \choose k}} = {1 \over 2}\,n\sum_{k = 0}^{n}{n \choose k}\color{red}{1}^{k} = {1 \over 2}\,n\left(1 + \color{red}{1}\right)^{n} \\[5mm] &\ \implies \bbox[10px,border:1px groove navy]{\bbox[10px,#ffd]{\sum_{k = 1}^{n}k{n \choose k}} = n\,2^{n - 1}} \end{align}

score 2 · Accepted Answer · answered Apr 24 '19 at 04:27

2

In each step you are saying that $\sum_{k=0}^{m}{n\choose k}=2^m$ which is not true! What is true is that $\sum_{k=0}^{m}{m\choose k}=2^m$. Be careful, note the difference.

The way to do it is using $k{n\choose k}=n{n-1\choose k-1}$. So at the end your sum is equal to $$\sum_{k=1}^nn{n-1\choose k-1}=n\sum_{j=0}^{n-1}{n-1\choose j}=n2^{n-1}$$

answered Apr 24 '19 at 04:27

Julian Mejia

4,458

The asker already checked out two threads. Your answer is from here, I think. – Jyrki Lahtonen Apr 24 '19 at 04:55

Summation of ${\sum_{k=1}^n k\binom{n}{k} = n2^{n-1}}$ using equations

2 Answers2